\^ 


t 


Digitized  by  the  Internet  Archive 

in  2008  with  funding  from 

Microsoft  Corporation 


http://www.archive.org/details/algebraschoolsOOnewcrich 


•MMOA  M3N 

^NVdWOO  DNiaNiaXOOa  QNV  ONIXNIUd 

S.MOUX 


•OO    2y    X10H    Aa^THH 

*t88X  '1881  ^xnoiUAcioo 


3  >5  V 


"^  IV^''"'^ 


SMODM.KK  isLOMIS     - 


Aa 


eaoa^^oo   a^v  st:oohos 


\, 


r"^ 


v»-^a-tDT:v 


'SaiEES   pVOIXVM^BXVH  S.SWOOAiSN 

\ 


PREFACE. 


The  course  of  algebra  embodied  in  the  present  work  is  substantially 
that  pursued  by  students  in  our  best  preparatory  and  scientific  schools 
and  colleges,  with  such  extensions  as  seemed  necessary  to  afford  an 
improved  basis  for  more  advanced  studies.  For  the  convenience  of 
;eacher3  the  work  is  divided  into  two  parts,  the  first  adapted  to  well- 
prepared  beginners  and  comprising  about  what  is  commonly  required  for 
admission  to  college ;  and  the  second  designed  for  the  more  advanced 
general  student.  As  the  work  deviates  in  several  points  from  the  models 
most  familiar  to  our  teachers,  a  statement  of  the  principles  on  which  it  is  ' 
constructed  may  be  deemed  appropriate. 

One  well-known  principle  underlying  the  acquisition  of  knowledge  is 
that  an  idea  cannot  be  fully  grasped  by  the  youthful  mind  unless  it  is 
presented  under  a  concrete  form.  Whenever  possible  an  abstract  idea 
rnust  be  embodied  in  some  visible  representation,  and  all  general  theorems 
must  be  presented  in  a  variety  of  special  forms  in  which  they  may  be 
seen  inductively.  In  accordance  with  this  principle,  numerical  exam- 
ples of  nearly  all  algebraic  operations  and  theorems  have  been  presented. 
For  the  purpose  of  illustration,  numbers  have  been  preferred  to  literal 
symbols  when  they  would  serve  the  purpose  equally  well.  The  relations 
of  positive  and  negative  algebraic  quantities  have  been  represented  by 
lines  and  directions  from  the  beginning  in  order  that  the  pupil  might  be 
able  to  give,  not  only  a  numerical,  but  a  visible,  meaning  to  all  algebraic 
quantities.  Should  it  appear  to  any  one  that  we  thus  detract  from  the 
generality  of  algebraic  quantities,  it  is  sufficient  to  reply  that  the  system 
is  the  same  which  mathematicians  use  to  assist  their  conceptions  of 
advanced  algebra,  and  without  which  they  would  never  have  been  able 
to  grasp  the  complicated  relations  of  imaginary  quantities.     Algebraic 

183978 


iv  ^  PREFACE. 

operations  with  pure  numbem  are  made  to  precede  the  use  of  symbols, 
and  the  latter  are  introduced  only  after  the  pupil  has  had  a  certain 
amount  of  familiarity  with  the  distinction  between  algebraic  and  numer- 
ical operations. 

Another,  but,  unfortunately,  a  less  familiar  fact  is,  that  all  mathematical 
conceptions  require  time  to  become  engrafted  upon  the  mind,  and  the 
more  time  the  greater  their  abstruseness.  It  is,  the  author  conceives, 
from  a  failure  to  take  account  of  this  fact,  rather  than  from  any  inherent 
defect  in  the  minds  of  our  youth,  that  we  are  to  attribute  the  backward 
state  of  mathematical  instruction  in  this  country,  as  compared  with  the 
continent  of  Europe.  Let  us  take  for  instance  the  case  of  the  student, 
commencing  the  calculus.  On  the  system  which  was  almost  universal 
among  us  a  few  years  ago,  and  which  is  still  widely  prevalent,  he  is  con- 
fronted at  the  outset  with  a  number  of  entirely  new  conceptions,  such 
as  those  of  variables,  functions,  increments,  infinitesimals  and  limits. 
In  his  first  lesson  he  finds  these  all  combined  with  a  notation  so  entirely 
diJOferent  from  that  to  which  he  has  been  accustomed,  that  before  the 
new  ideas  and  forms  of  thought  can  take  permanent  root  in  his  mind, 
he  is  through  with  the  subject,  and  all  that  he  has  learned  is  apt  to  vanish 
from  his  memory  in  a  few  months. 

The  author  conceives  that  the  true  method  of  meeting  this  difficulty  is 
to  adopt  the  French  and  Grerman  plan  of  teaching  algebra  in  a  broader 
way,  and  of  introducing  the  more  advanced  conceptions  at  the  earhest 
practicable  period  in  the  course.  Accordingly,  the  attempt  is  made  in  the 
present  work  to  introduce  each  advanced  conception,  disguised  perhaps 
under  some  simple  form,  in  advance  of  its  general  enunciation  and  at  as 
early  a  period  as  the  student  can  be  expected  to  understand  it.  In  doing 
this,  logical  order  is  frequently  sacrificed  to  the  exigencies  of  the  case, 
because  there  are  several  subjects  with  which  a  certain  amount  of  famil- 
iarity must  be  acquired  before  the  pupil  can  even  clearly  comprehend 
general  statements  respecting  them. 

A  third  feature  of  the  work  is  that  of  subdividing  each  subject  as 
minutely  as  possible,  and  exercising  the  pupil  on  the  details  preparatory  to 
combining  them  into  a  whole.  To  cito  one  or  two  instances :  a  difficulty 
which  not  only  the  beginner  but  the  expert  mathematician  frequently 
meets  is  that  of  stating  his  conceptions  in  algebraic  language.  Exercises 
in  such  statements  have  therefore  been  made  to  precede  any  solution  of 


PREFACE,  V 

problems.  In  general  each  principle  whicli  is  to  be  presented  or  used  is 
stated  singly,  and  tbe  pupil  is  practiced  upon  it  before  proceeding  to 
another. 

Subjects  have  for  the  most  part  been  omitted  which  do  not  find  appli- 
cation either  in  the  work  itself  or  in  subsequent  parts  of  the  usual  course 
of  mathematics,  or  which  do  not  conduce  to  a  mathematical  training. 
Sturm's  Theorem  has  been  entirely  omitted,  and  a  more  simple  process 
substituted.  The  subject  of  the  greatest  common  divisor  of  two  polyno- 
mials has  been  postponed  to  what  the  author  considers  its  proper  place, 
in  the  genaral  theory  of  equations.  It  has,  however,  been  presented  in 
Buch  a  form  that  it  can  be  taught  to  pupils  preparing  for  colleges  where 
it  is  still  required  for  admission. 

Thoroughness  at  each  step  has  been  aimed  at  rather  than  multiplicity 
of  subjects.  It  is,  the  author  conceives,  a  great  and  too  common 
mistake  to  present  a  mathematical  subject  to  the  mind  of  the  student 
wi^out  sufficient  fulness  of  explanation  and  variety  of  illustration  to 
enable  him  to  comprehend  and  apply  it.  If  he  has  not  time  to  master  a 
complete  course,  it  is  better  to  omit  entirely  what  is  least  necessary  than 
to  gain  time  by  going  rapidly  over  a  great  number  of  things.  Some 
hints  to  those  who  may  not  have  time  to  master  the  whole  work  may 
therefore  be  acceptable. 

Part  I  is  essential  to  every  one  desiring  to  make  use  of  algebra.  Book 
YIII,  especially  the  concluding  sections  on  notation,  is  to  be  thoroughly 
mastered,  before  going  farther,  as  forming  the  foundation  of  advanced 
algebra ;  and  affording  a  very  easy  and  valuable  discipline  in  the  language 
of  mathematics.  Afterward,  a  selection  may  be  made  according  to  cir- 
cumstances. The  student  who  is  pursuing  the  subject  for  the  sole 
purpose  of  liberal  training,  and  without  intending  to  advance  beyond  it, 
will  find  the  theories  of  numbers  and  the  combinatoiy  analysis  most 
worthy  of  study.  The  theory  of  probabilities  and  the  method  in  which 
it  is  applied  to  such  practical  questions  as  those  connected  with  insurance 
will  be  of  especial  value  in  training  his  judgment  to  the  affairs  of  life. 

The  student  who  intends  to  take  a  full  course  of  mathematics  with  a 
view  of  its  application  to  physics,  engineering,  or  other  subjects,  may,  if 
necessary,  omit  the  book  on  the  theory  of  numbers,  and  portions  of  the 
chapter  on  the  summation  of  series.  Functions  and  the  functional  notation, 
the  doctrine  of  limits,  and  the  general  theory  of  equations  will  '  'iim  his 


VI  {PREFACE. 

especial  attention,  while  the  theory  of  imaginary  quantities  will  be  studied 
mainly  to  secure  thoroughness  in  subsequent  parts  of  hiv  course. 

As  it  has  frequently  been  a  part  of  the  author's  duty  to  ascertain  what 
is  really  left  of  a  course  of  mathematical  study  in  the  minds  of  those 
who  have  been  through  college,  some  hints  on  the  best  methods  of 
study  in  connection  with  the  present  work  may  be  excused.  If  asked 
to  point  out  the  greatest  error  in  our  usual  system  of  mathematical 
instruction  from  the  common  school  upward,  he  would  reply  that  it  con- 
sisted in  expending  too  much  of  the  mental  power  of  the  student  upon 
problems  and  exercises  above  his  capacity.  With  the  exception  of  the 
fundamental  routine- operations,  problems  and  exercises  should  be  confined 
to  insuring  a  proper  understanding  of  the  principles  involved :  this  once 
ascertained,  it  is  better  that  the  student  should  go  on  rather  than  expend 
time  in  doing  what  it  is  certain  he  can  do.  Problems  of  some  difficulty 
are  found  among  the  exercises  of  the  present  work;  they  are  inserted 
rather  to  give  the  teacher  a  good  choice  from  which  to  select  than  to 
require  that  any  student  should  do  them  all. 

It  would,  the  author  conceives,  be  found  an  improvement  on  our  usual 
system  of  teaching  algebra  and  geometry  successively  if  the  analytic  and 
the  geometric  courses  of  mathematics  were  pursued  simultaneously.  The 
former  would  include  algebra  and  the  calculus,  the  latter  elementary 
geometry,  trigonometry,  and  analytic  geometry.  The  analytic  course 
would  then  furnish  methods  for  the  geometric  one,  and  the  latter  would 
fiirnish  applications  and  illustrations  for  the  analytic  one. 

The  Key  to  the  work  contains  not  only  the  usual  solutions,  but  the 
explanations  and  demonstrations  of  the  less  familiar  theorems,  and  a 
number  of  additional  problems. 

The  author  desires,  in  conclusion,  to  express  his  obligation  to  the  many 
friends  who  have  given  him  suggestions  respecting  the  work,  and  espe- 
cially to  Professor  J.  Howard  Gore,  of  the  Columbian  University,  who 
has  furnished  solutions  to  most  of  the  problems,  and  given  the  benefit  of 
his  experience  on  many  points  of  detail. 


Note. — Answers  to  exercises,  requiring  calculation  or  written  worTc^  are 
published  separately  in  pamphlet  form,  and  will  be  supplied  without 
cha1;ge  when  applied  for  by  teachers. 


TABLE   OF    CONTENTS, 


PAET    L— ELEMENTAKY    COURSE. 

BOOK  I.— THE  ALGEBRAIC  LANGUAGE, 

Chapter  I.— Algebraic  Numbers  and  Operations,  3,  Genr  ^l 
Definitions,  3.  Algebraic  Numbers,  4,  Algebraic  Addition,  6. 
Subtraction,  8.    Multiplication,  9.    Division,  11. 

Chapter  II.— Algebraic  Symbols,  12.  Symbols  of  Quantity,  12. 
Signs  of  Operation,  13. 

Chapter  III.— Formation  of  Compound  Expressions,  17,  Funda- 
mental Principles,  17.    Definitions,  19. 

Chapter  IV. — Construction  of  Algebraic  Expressions,  22,  Exer- 
cises in  Algebraic  Language,  25. 

BOOK  IL— ALGEBRAIC  OPERATIONS. 

General  Remarks,  28.     Definitions,  28. 

Chapter  I. — Algebraic  Addition  and  Subtraction,  30.  Algebraic 
Addition,  30.  Algebraic  Subtraction,  33.  Clearing  of  Parenthe- 
ses, 35.    Compound  Parentheses,  37. 

Chapter  II,— Multiplication,  38.  General  Laws  of  Multiplication,  38. 
Multiplication  of  Positive  Monomials,  40.  Rule  of  Signs  in 
Multiplication,  41.  Products  of  Polynomials  by  Monomials,  44 
Multiplication  of  Polynomials  by  Polynomials,  47. 

Chapter  'ill. — Division,  52.  Division  of  Monomials  by  Monomials,  52. 
Rule  of  Signs  in  Division,  53.    Division  of  Polynomials  by  Mono- 


Viii  '    GDNTENT8. 

\ 
mials,  54.    Factors  and  Multiples,  55.    Factors  of  Binomials,  58. 

Least  Common  Multiple,   61.     Division  of   one  Polynomial  by 

another,  62. 

Chapter  IV.— Of  Algebraic  Fractions,  67.  JSegative  Exponents,  71. 
Dissection  of  Fractions,  73.  Aggregation  of  Fractions,  74.  Factor- 
ing Fractions,  78.  Multiplication  and  Division  of  Fractions,  79. 
Division  of  one  Fraction  by  another,  82.  Reciprocal  Relations  ol 
Multiplication  and  Division,  83. 


BOOK  III.— OF  EQUATIONS. 

Chapter,  I.— The  Reduction  of  Equations,  85.  Axioms,  87.  Opera- 
tions of  Addition  and  Subtraction — transposing  Terms,  87. 
Operation  of  Multiplication,  89.  Reduction  to  the  Normal  Form, 
90.     Degree  of  Equations,  93. 

Chapter  II. — Equations  of  the  First  Degree  with  One  Un- 
known Quantity,  94.  Problems  leading  to  Simple  Equations,  99. 
Problem  of  the  Couriers,  105.    Problems  of  Circular  Motion,  108. 

Chapter  III.— Equations  of  the  First  Degree  with  Several 
Unknown  Quantities,  109.  Equations  with  Two  Unknown 
Quantities,  109,  Solution  of  a  pair  of  Simultaneous  Equations 
containing  Two  Unknown  Quantities,  109.  Elimination  by  Com- 
parison, 110.  Elimination  by  Substitution,  111,  Elimination  by 
Addition  or  Subtraction,  112.  Problem  of  the  Sum  and  Differ- 
ences, 113.  Equations  of  the  First  Degree  with  Three  or  More 
Unknown  Quantities,  116.  Elimination,  116.  Equivalent  and 
Inconsistent  Equations,  121. 

Chapter  IV. — Of  Inequalities,  123. 


BOOK  IV.— RATIO  AND  PROPORTION. 
Chapter  I. — Nature  of  a  Ratio,  128.    Properties  of  Ratios,  132. 

Chapter  II. — Proportion,  133.     Theorems  of  Proportion,  134.     The 
Mean  Proportional,  138.    Multiple  Proportions,  139.     ^ 


contents:  IX 


BOOK  v.— OF  POWERS  AND   ROOTS. 

Chapter  I. — Involution,  144.  Involution  of  Products  and  Quotients, 
144.  Involution  of  Powers,  145.  Case  of  Negative  Exponents,  147. 
Algebraic  Sign  of  Powers,  148.  Involution  of  Binomials — tlie 
Binomial  Theorem,  148.    Square  of  a  Polynomial,  153. 

Chapter  II.— Evolution  and  Fractional  Exponents,  155.  Powers 
of  Expressions  with  Fractional  Exponents,  157. 

Chapter  III. — Reduction  of  Irrational  Expressions,  159.  Defini- 
tions, 159.  Aggregation  of  Similar  Terms,  160.  Factoring  Surds, 
161.  Perfect  Squares,  166.  To  Complete  the  Square,  167.  Irra- 
tional Factors,  169. 


BOOK  VI.— EQUATIONS  REQUIRING  IRRATIONAL  OPERATIONS. 

Chapter  I.— Equations  with  Two  Terms  only,  170.  Solution  of  a 
Binomial  Equation,  170.  Special  Forms  of  Binomial  Equations, 
171.    Positive  and  Negative  Roots,  172. 

Chapter  II. — Quadratic  Equations,  174.  Solution  of  a  Complete 
Quadratic  Equation,  175.  Equations  which  may  be  reduced  to 
Quadratics,  179.  Factoring  a  Quadratic  Equation,  184.  Equations 
having  Imaginary  Roots,  188. 

Chapter  III. — Reduction  of  Irrational  Equations  to  the  Normal 
Form,  189.     Clearing  of  Surds,  189. 

Chapter  IV. — Simultaneous  Quadratic  Equations,  193. 


BOOK  VII.— PROGRESSIONS. 

Chapter  I. — Arithmetical  Progression,  200.  Problems  in  Pro- 
gression, 202. 

Chapter  II.— Geometrical  Progression,  207.  Problems  of  Geo- 
metrical Progression,  208.  Limit  of  the  Sum  of  a  Progression, 
^M     Compound  Interest,  217. 


X  CONTENTS. 

PART    II.— ADVANCED    COUESE. 

BOOK  VIII.— RELATIONS  BETWEEN  ALGEBRAIC  QUANTITIES. 

Functions  and  their  Notation,  221.  Equations  of  tlie  First  Degree 
between  Two  Variables,  224.  Notation  of  Functions,  230.  Func- 
tions of  Several  Variables,  282.  Use  of  Indices,  233.  Miscellaneous 
Functions  of  Numbers,  235. 

BOOK  IX.— THE  THEORY  OF   NUMBERS. 

Cttapter  I.— The  Divisibility  of  Numbers,  238.  Division  into  Prime 
Factors,  239.  Common  Divisors  of  two  numbers,  240.  Relations 
of  numbers  to  their  Digits,  245.  Divisibility  of  Numbers  and  their 
Digits-,  245.  Prime  Factors  of  Numbers,  248.  Elementary  Theorems, 
251.     Binomial  Coefficients,  251.     Divisors  of  a  Number,  254. 

Chapter  II.— Of  Continued  Fractions,  258.  Relations  of  Converging 
Fractions,  267.     Periodic  Continued  Fractions,  270. 

BOOK  X.— THE  COMBINATORY  ANALYSIS. 

Chapter  I. — Permutations,  273.  Permutation  of  Sets,  275.  Circular 
Permutations,  277.  Permutations  when  Several  of  the  Things 
are  Identical,  279.  The  two  classes  of  Permutations,  281.  Sym- 
metric Functions,  284. 

Chapter  II. — Combinations,  285.  Combinations  with  Repetition, 
287.  Speeial  Cases  of  Combinations,  289.  The  Binomial  Theorem 
when  the  Power  is  a  Whole  Number,  296. 

Chapter  III.— Theory  of  Probabilities,  299.  Probabilities  depend- 
ing upon  Combinations,  300.  Compound  Events,  305.  Cases  of 
Unequal  Probability,  310,  Application  to  Life  Insurance,  316. 
Table  of  Mortality,  318. 

BOOK  XI.— OF  SERIES  AND  THE  DOCTRINE  OF  LIMITS. 
Chapter  I.-— Nature  of  a  Series,  321.    Notation  of  Sums,  32-     < 


CONTENTS.  Xi 

Chapter  IL— Development  in  Powers  op  a  Variable,  326. 
Method  of  Indeterminate  Coefficients,  327.  Multiplication  of  Two 
Infinite  Series,  333. 

Chapter  III.— Summation  op  Series.  Of  Figurate  Numbers,  336. 
Enumeration  of  Triangular  Piles  of  Shot,  339.  Sum  of  the 
Similar  Powers  of  an  Arithmetical  Progression,  341.  Other  Series, 
345.    Of  Differences,  350.    Theorems  of  Differences,  355. 

Chapter  IV. — The  Doctrine  of  Limits,  358.  Notation  of  the 
Method  of  Limits,  361.     Properties  of  Limits,  364. 

y 

Chapter  V. — The  Binomial  and  Exponential  Theorems.  The 
Binomial  Theorem  for  all  values  of  the  Exponent,  368.  The 
Exponential  Theorem,  373. 

Chapter  VI.— Logarithms,  378.  Properties  of  Logarithms,  378.  Com- 
parison of  Two  Systems  of  Logarithms,  384. 

BOOK  XII.— IMAGINARY  QUANTITIES. 

Chapter  I.— Operations  with  the  Imaginary  Unit,  391.  Addi- 
tion of  Imaginary  Expressions,  393.  Multiplication  of  Imaginary 
Quantities,  393.  Reduction  of  Functions  of  i  to  the  Normal 
Form,  396. 

Chapter  II.— The  Geometrical  Representation  op  Imaginary 
Quantities,  404. 

BOOK  XIIL— THE  GENERAL  THEORY  OF  EQUATIONS. 

Every  Equation  has  a  Root,  416.  Number  of  Roots  of  General 
Equation,  418.  Relations  between  Coefficients  and  Roots,  425. 
Derived  Functions,  427.  Significance  of  the  Derived  Function,  430. 
Forms  of  the  Roots  of  Equation,  43 1.  Decomposition  of  Rational 
Fractions,  433.  Greatest  Common  Divisor  of  Two  Functions,  438. 
Transformation  of  Equations,  442.  Resolution  of  Numerical  Equa- 
tions, 447. 


^       or  THE  \ 

UNIVERSITY  |. 

OF 


BOOK    I. 
TUB     ALGEBRAIC     LANGUAGE. 


CHAPTER    I. 

OF    ALGEBRAIC    NUMBERS    AND    OPERATIONS. 


General   Definitions. 

1.  Definition.  Mathematics  is  the  science  which 
treats  of  the  relations  of  magnitudes. 

The  magnitudes  of  mathematics  are  time,  space,  force, 
value,  or  other  things  which  can  be  thought  of  as  entirely 
made  up  of  parts. 

3.  Def.  A  Quantity  is  a  definite  portion  of  any 
magnitude. 

Example.  Any  definite  number  of  feet,  miles,  acres, 
bushels,  years,  pounds,  or  dollars,  is  a  quantity. 

3.  Def.  Algebra  treats  of  those  relations  which 
are  true  of  quantities  of  every  kind  of  magnitude. 

4.  The  relations  treated  of  in  Algebra  are  discovered 
by  means  of  numbers. 

To  measure  a  quantity  by  number,  we  take  a  certain  por- 
>n  of  the  magnitude  to  be  measured  as  a  unit,  and  express 
ow  many  of  the  units  the  quantity  contains. 
Eemark.     It  is  obviously  essential  that  the  quantity  and 
its  unit  shall  be  the  same  kind  of  magnitude. 

o.  Def,  A  Concrete  Number  is  one  in  which  the 
kind  of  quantity  which  it  measures  is  expressed  or 
understood  ;  as  7  miles ^  3  days,,  or  10  'pounds. 


4  THE  ALGEBRAIC   LANGUAGE. 

6.  Def.  An  Abstract  Number  is  one  in  wiiii^ix  Uv/ 
particular  kind  of  unit  is  expressed  ;  as  7,  3,  or  10. 

Eemark.  An  abstract  number  may  be  considered  as  a 
concrete  one  expressing  a  certain  number  of  units,  without 
respect  to  the  kind  of  units.     Thus,  7  means  7  units. 

Algebraic  Numbers. 

7.  In  Arithmetic,  the  numbers  begin  at  0,  and  in- 
crease without  limit,  as  0,  1,  2,  3,  4,  etc.  But  the 
quantities  we  usually  measure  by  numbers,  as  time 
and  space,  do  not  really  begin  at  any  point,  but  extend 
without  end  in  opposite  directions. 

For  example,  time  has  no  beginning  and  no  end.  An 
epoch  of  time  1000  years  from  Christ  may  be  either  1000  years 
after  Christ,  or  1000  years  before  Christ. 

A  heavy  body  tends  to  fall  to  the  ground.  A  body  which 
did  not  tend  to  move  at  all  when  unsupported  would  have  no 
weight,  or  its  weight  would  be  0.  If  it  tended  to  rise  upward, 
like  a  balloon,  it  would  have  the  opposite  of  weight. 

If  we  have  to  measure  a  distance  from  any  point  on  a 
straight  line,  we  may  measure  out  in  either  direction  on  the 
line.     If  the  one  direction  is  east,  the  other  will  be  west. 

One  who  measures  his  wealth  is  poorer  by  all  that  he  owes. 
If  he  owes  more  than  he  possesses,  he  is  worth  less  than 
nothing,  and  there  is  no  limit  to  the  amount  he  may  owe. 

8.  In  order  to  measure  such  quantities  on  a  uni- 
form system,  the  numbers  of  Algebra  are  considered  as 
increasing  from  0  in  two  opposite  directions.  Those  in 
one  direction  are  called  Positive;  those  in  the  other 
direction  Negative. 

9.  Positive  numbers  are  distinguished  by  the  sign 
+  ,  plus  ;  negative  ones  by  the  sign  — ,  minus. 

If  a  positive  number  measures  years  after  Christ,  a  negative 
one  will  mean  years  before  Christ. 

If  a  positive  number  is  used  to  measure  toward  the  right,  a 
negative  one  will  measure  toward  the  left. 


ALGEBRAIC  NUMBERS.  5 

If  a  positive  number  measures  weight,  the  negative  one 
will  imply  levity,  or  tendency  to  rise  from  the  earth. 

If  a  positive  number  measures  property,  or  credit,  the  nega- 
tive one  will  imply  debt. 

10,  TBe  series  of  algebraic  numbers  will  therefore 
I  be  considered  as  arranged  in  the  following  way,  the 

series  going  out  to  infinity  in  both  directions. 

-=©91    NEGATIVE   DIRECTION.  POSITIVE   DIRECTION.     ^T 

Before.  After. 

Downward,  Upward. 

Debt.  Credit, 

etc.  etc. 

etc.  -5,  -4,  -3,  -2,  -1,  0,  +1,  +2,  +3,  +4,  +5,  etc. 

Rem.  It  matters  not  which  direction  we  take  as  the 
positive  one,  so  long  as  we  take  the  opposite  one  as 
negative. 

If  we  take  time  before  as  positive,  time  after  will  be  nega- 
tive ;  if  we  take  west  as  the  positive  direction,  east  will  be 
negative;  if  we  take  debt  as  positive,  credit  will  be  negative. 

11.  Positive  and  negative  numbers  may  be  conceived 
as  measuring  distances  from  a  fixed  point  on  a  straight 
line,  extending  indefinitely  in  both  directions,  the  dis- 
tances one  way  being  positive,  and  the  other  way 
negative,  as  in  the  following  scheme :  * 

etc.    -7,  -6.  -5,  -4,  -3,  -2,  -1,     0,  +1,  +2,  +3,  +4,  +5,  +6,  +7.  etc. 

I       I       I       I       I       I       I       I       I       I        I        I       I       i       I 

In  this  scale,  the  distance  between  any  two  consecu- 
tive numbers  is  considered  a  unit  or  unit  step. 

13.  Def.  The  signs  +  and  —  are  called  the  Alge- 
braic Signs,  because  they  mark  the  direction  in  which 
the  numbers  following  them  are  to  be  taken. 

*  The  student  should  copy  this  scale  of  numbers,  and  have  it  before 

^im  in  studying  tlio  present  chapter. 


6  THE  ALGEBRAIC  LANGUAGE. 

The  sign  +  may  be  omitted  before  positive  numbers,  when 
no  ambiguity  is  thus  produced.  The  numbers  2,  5,  12,  taken 
alone,  signify  +2,  +5,  +12.  But  the  negative  sign  must 
always  be  written  when  a  negative  number  is  intended. 

13.  Def,  One  number  is  said  to  be  Algebraically- 
Greater  than  another  when  on  the  preceding  scale  it 
lies  to  the  positive  Cright  hand)  side.     Thus, 

—  2    is  algebraically  greater  than    —  7 ; 
0     "  "  "  "         _2; 

5     "  "  "  "        —  5. 

Alg^ebraic  Addition. 

14.  Def.  In  Algebra,  Addition  means  the  combi- 
nation of  quantities  according  to  their  algebraic  signs, 
the  positive  quantities  being  counted  one  way  or  added, 
and  negative  ones  the  opposite  way  or  subtracted. 

15.  Def.  The  Algebraic  Sum  of  several  quantities 
is  the  surplus  of  the  positive  quantities  over  the  nega- 
tive ones,  or  of  the  negative  quantities  over  the  positive 
ones,  according  as  the  one  or  the  other  is  the  greater. 

The  sum  has  the  same  algebraic  sign  as  the  prepon- 
derating quantity. 

Example.    The  sum  of 

+  ^     and  —  7  is         0 ; 

+  9       ''  -^7  ^^  4-2; 

4-5       ^^  -_7  ^^  _2. 

The  sum  of  several  positive  numbers  may  be  represented 
on  the  line  of  numbers,  §  11,  by  the  length  of  the  line  formed 
by  placing  the  lengths  represented  by  the  several  numbers 
end  to  end.  The  total  length  will  be  the  sum  of  the  partial 
lengths. 

If  any  of  the  numbers  are  negative,  the  algebraic  sum  is 
represented  by  laying  their  lengths  off  in  the  opposite  direction. 

Example  1.  The  algebraic  sum  of  the  four  numbers  9, 
—  7,  1,   —6,  would  be  represented  thus : 


algebbaIg  addition. 


ET 


Here,  starting  from  0,  we  measure  9  to  the  right,  then  7 
to  the  left,  then  1  to  the  right,  then  6  to  the  left.  The  result 
would  be  3  steps  to  the  left  from  0,  that  is,  —  3.  Thus,  —  3 
is  the  algebraic  sum  of  +9,  —7,  +1,  and  —6. 

Ex.  2.  If  we  imagine  a  person  to  walk  back  and  forth 
along  the  line  of  numbers,  his  distance  from  the  starting- 
point  will  always  be  the  algebraic  sum  of  the  separate  distances 
he  has  walked. 

Ex.  3.  A  man's  wealth  is  the  algebraic  sum  of  his  posses- 
sions and  credits,  the  debts  which  he  owes  being  negative 
credits*.  If  he  has  in  money  $1000,  due  from  A  $1200,  due  to 
X  $500,  due  to  Y  $350,  his  possessions  would,  in  the  language 
of  algebra,  be  summed  up  as  follows  : 

Cash,  ,  .  ...  +  $1000 

Due  from  A,  .  .  .        _  +1200 

Due  from  X,  .  -  .        .  —      500 

Due  from  Y,  ,.  .  .        .  —      350 

Sum  total,  .  .  .        .  +  $1350 

[In  the  language  of  Algebra,  the  fact  that  he  owes  X  $500 
may  be  expressed  by  saying  that  X  owes  him  —  $500.] 

16.  Bef.  To  distinguish  between  ordinary  and 
algebraic  addition,  the  former  is  called  Numerical  or 
Arithmetical  addition. 

Hence,  the  numerical  sum  of  several  numbers 
means  their  sum  as  in  arithmetic,  without  regard  to 
their  signs. 

17.  Rem.  In  Algebra,  whenever  the  word  sum 
is  used  without  an  adjective,  the  algebraic  sum  is 
understood. 


8  THE  ALOEBBAJg   LANGUAGE. 

Algebraic    Subtraction. 

18.  Memorandum  of  arithmetical  defiyiitions  and oijerations. 

The  Subtrahend  is  the  quantity  to  be  subtracted. 

The  Minuend  is  the  quantity  from  which  the  subtrahend 
is  taken. 

The  Remainder  or  Difference  is  what  is  left. 

If  we  subtract  4  from  7,  the  remainder  3  is  the  number  of 

anit  steps  on  the  scale  of  numbers  (§  11)  from  +4  to  +7. 

This  is  true  of  any  arithmetical  difference  of  numbers.     In 

Algebra,  the  operation  is  generalized  as  follows : 

19.  Def.  The  Algebraic  Difference  of  two  num- 
bers is  represented  by  the  distance  from  one  to  the 
other  on  the  scale  of  numbers. 

The  number  from  which  we  measure  is  the  Subtra- 
hend. 

That  to  which,  we  measure  is  the  Minuend. 

If  the  minuend  is  algebraically  the  greater  (§  13), 
the  difference  is  positive. 

If  the  minuend  is  less  than  the  subtrahend,  the  dif- 
ference is  negative. 

In  Arithmetic  we  cannot  subtract  a  greater  number  from  a 
less  one.  But  there  is  no  such  restriction  in  Algebra,  because 
algebraic  subtraction  does  not  mean  taking  away,  but  finding 
a  difference.  However  the  minuend  and  subtrahend  may  be 
situated  on  the  scale,  a  certain  number  of  spaces  toward  the 
right  or  toward  the  left  will  always  carry  us  from  the  subtra- 
hend to  the  minuend,  and  these  spaces  make  up  the  difference 
of  the  two  numbers. 

30.  The  general  rule  for  algebraic  subtraction  may  be 
deduced  as  follows :  It  is  evident  that  if  we  pass  from  the 
subtrahend  to  0  on  the  scale,  and  then  from  0  to  the  minuend, 
the  algebraic  sum  of  these  two  motions  will  be  the  entire  space 
between  the  subtrahend  and  minuend,  and  will  therefore  be 
the  remainder  required.  But  the  first  motion  will  be  equal  to 
the  subtrahend,  but  positive  if  that  quantity  is  negative,  and 
vice  versa,  and  the  second  motion  will  be  equal  to  the  minuend. 


ALGEBRAIC  MULTIPLICATION.  9 

Hence  the  remainder  will  be  found  by  changing  the  algebraic 
sign  of  the  subtrahend^  and  then  adding  it  algebraically  to  the 
minuend. 

EXAMPLES. 

Subtracting   +  5  from  +  8,  the  difference  is      8  —  5  =  3. 

+  8     ^^      +  5,    "  "  "      5  —  8  =  —  3. 

+  8     ^^     —  5,    ^^  "  ^^_5_8  =  — 13. 

—  8     ^^  h,    "  "         ^^      5  H   8  =  +  13. 

+  13     "  0,    "  "    -^  "        ^  —13. 

_  13     ''  0,    "  "      ^  ^'  _l_  13. 

31.  By  comparing  algebraic  addition  and  subtraction,  it 
will  be  seen  that  to  subtract  a  positive  number  ig* the  same 
thing  as  to  add  its  negative,  and  vice  versa.     Thus, 

To  subtract  5  from  8  gives  the  same  result  as  to  add  —  5 
to  8,  namely  3. 

To  subtract  —  5  from  8  gives  8  +  5,  namely  13. 

Hence,  algebraic  subtraction  is  equivalent  to  the 
algebraic  addition  of  a  number  with  the  opposite 
algebraic  sign.  Algebraists,  therefore,  do  not  consider 
subtraction  as  an  operation  distinct  from  addition. 

Algebraic   Multiplication. 

33,  Memorandum  of  arithmetical  definitions. 

The  Multiplicand  is  the  quantity  to  be  multiplied. 
The  Multiplier  is  the  number  by  which  it  is  multiplied. 
The  result  is  called  the  Product. 

Factors  of  a  number  are  the  multiplicand  and  multiplier 
which  produce  it. 

23.  To  multiply  any  algebraic  quantity  by  a  posi- 
tive whole  number  means,  as  in  Arithmetic,  to  take  it  a 
number  of  times  equal  to  the  multiplier. 

Thus,'  4x3=       4  +  4  +  4=+ 12; 

—  4x3=— 4  —  4  —  4=  —  12. 

The  product  of  a  negative  multiplicand  by  a  positive 
multiplier  will  therefore  be  negative. 


10  THE  ALGEBRAIC  LANGVAOE. 

34.  If  the  multiplier  is  negative,  the  sign  of  the 
product  will  be  the  opposite  of  what  it  would  be  if  the 
multiplier  were  positive. 

Thus,  +4  X  —3  =  —12; 

-4  X  -3  =   +  12. 
The  product  of  two  negative  factors  is  therefore 
positive. 

35.  The  most  simple  way  of  mastering  the  use  of  algebraic 
signs  in  multiplication  is  to  think  of  the  sign  —  as  meaning 
opposite  in  direction.  Thus,  in  §11,  — 4  is  opposite  in 
direction  to  +  4,  the  direction  being  that  from  0.  If  we  mul- 
tiply this  negative  factor  by  a  negative  multiplier,  the  direction 
will  be  the  opposite  of  negative,  that  is,  it  will  be  positive,  A 
third  negative  factor  will  make  the  product  negative  again,  a 
fourth  one  positive,  and  so  on.    For  example, 

-3  X  -4=  +12; 
—2  X  —3  X  —4  =  — 2  X  +12  =  —24; 
—3  X  —2  X  -3  X  -4  rrz  -3  X  -24  =  +  72 ; 
etc.  etc. 

Hence, 

36.  Theorem,  The  continued  product  of  an  even 
number  of  negative  factors  is  positive  ;  of  an  odd  num- 
ber, negative. 

Rem.  Multiplying  a  number  by  —1  simply  changes 
its  sign. 

Thus,  _|-4  X  — 1  =  —  4; 

-4  X  -1  =  +  4. 

EXERCISES. 

Find  the  algebraic  sums  of  the  following  quantities  : 

1.  4  —  6  +  12  —  1—18. 

2.  —  6  —  3  —  8. 

3.  _  6  —  10  —  9  +  34. 

4.  Subtract  the  sum  in  Ex.  3  from  the  sum  in  Ex.  2. 

5.  Subtract  the  sum  5  —  6  +  3—1  —  16,  from  the  sum 
—  2  —  7-4  +  8. 


ALGEBRAIC   DIVISION.  11 

6.  Subtract  the  sum  5  —  6+3-— 1  —  16,  from  the  sum 
7  _  3  _  8  +  4. 

7.  Form  the  product  —7x8. 

8.  Form  the  product  —8x7. 

9.  Form  the  product  6x— 5x7x  —  4. 

10.  Form  the  product  — 6x—  llx8x—  2. 

11.  Form  the  product  — Ix— -Ix— Ix—  1. 

1 2.  Subtract  the  sum  in  Ex.  1  from  the  sum  in  Ex.  3,  and 
multiply  the  remainder  by  the  sum  in  Ex.  2. 

13.  Subtract  8  from  —  3,   —  3  from  -1,-1  from  8,  and 
find  the  sum  of  the  three  remainders. 

14.  Subtract  7  from  —  9  and  the  remainder  from  2,  and 
multiply  the  result  by  the  product  in  Ex.  7. 

Algfebraic   Division. 

37.  Memorandum  of  aritlimetical  definitions. 

The  Dividend  is  the  quantity  to  be  divided. 
The  Divisor  is  the  number  by  which  it  is  divided. 
The  Quotient  is  the  result. 

38.  Bule  of  Signs  in  Division,  The  requirement 
of  division  in  Algebra  is  the  same  as  in  Arithmetic  ; 
namely. 

The  product  of  the  quotient  hy  the  divisor  must  he 
equal  to  the  dividend. 

In  Algebra,  two  quantities  are  not  equal  unless  they  have 
the  same  algebraic  sign.     Therefore  the  product, 

quotient  x  divisor 

must  have  the  same  algebraic  sign  as  the  dividend.     From 
this  we  can  deduce  the  rule  of  signs  in  division. 

Let  us  divide  6  by  2,  giving  6  and  2  both  algebraic  signs, 
and  find  the  signs  of  the  quotient  3 : 

+  3x+2=+6;  therefore,  +6  divided  by  +2  gives  -i-3. 

+  3x— 2=— 6;  ''  —6       "        ''—2     ''      +3. 

— 3x+2=:— 6;  "  —6       ''        ''   -\-2     ''      —3. 

— 3  X  — 2  z=  +6;  "  +6       "        "  —2     "     —3. 


12  THE  ALGEBRAIC  LANGUAGE. 

Hence,  the  rule  of  signs  is  tlie  same  in  division  as  in  mul- 
tiplication, namely : 

Like  signs  in  dividend  and  divisor  give  + .     Unlike 
signs  give  — . 

EXERCISES. 

Execute  the  following  algebraic  divisions,  expressing  each 
result  as  a  whole  number  or  vulgar  fraction : 

1.  Dividend,  —  7  +  10  —  11  +  25  ;  divisor,  20  —  3. 

2.  Dividend,  12  —  3  +  15  —  10  ;       divisor,  3  —  10. 

3.  Dividend,  25  —  36  +  6  —  20  ;       divisor,  —3  +  8. 

4.  Dividend,  —  7  x  —  8  ;  divisor,  —8  +  4. 

5.  Dividend,  56  +  8  x  —  3  ;  divisor,  —  4  —  4. 

6.  Dividend,  —  24  x  —  1 ;  divisor,  —  3  x  —  3. 

7.  Dividend,  —13  x  —10  x  —  8  ;  divisor,  — 4x5x— 6. 

8.  Dividend,  —  1  x  —  1 ;  divisor,  —  3  x  —  3. 


CHAPTER    II. 
ALGEBRAIC     SYM  BOLS. 


Symbols  of  Quantity. 

39.  Algebraic  quantities  may  be  represented  by 
letters  of  the  alphabet,  or  other  characters.  1 

The  characters  of  Algebra  are  called  Symbols. 

30.  Def.  The  Value  of  an  algebraic  symbol  is  the 
quantity  which  it  represents  or  to  which  it  is  equal. 

The  value  of  a  symbol  may  be  any  algebraic  quan- 
tity whatever,  positive  or  negative,  which  we  choose  to 
assign  to  the  symbol. 

31.  The  language  of  Algebra  differs  in  one  respect  from 
ordinary  language.     In  the  latter,  each  special  word  or  sign 


SIGJVS    OF   OPERATION.  13 

has  a  definite  and  invariable  meaning,  which  every  one  who 
uses  the  language  must  learn  once  for  all.  But  in  Algebra  a 
symbol  may  stand  for  any  quantity  which  the  writer  or  speaker 
chooses,  and  his  results  must  be  interpreted  according  to  this 
meaning. 

33.  The  same  character  may  be  used  to  represent  several 
quantities  by  applying  accents  or  attaching  numbers  to  it  to 
distinguish  the  different  quantities.  Thus,  the  four  symbols, 
a,  a',  a",  a'",  may  represent  four  different  quantities.  The 
symbols  ai,  a.2,  ^3,  «4?  «5?  ^tc,  maybe  used  to  designate  any 
number  of  quantities  which  are  distinguished  by  the  small 
number  written  after  the  letter  a. 

Si^iis  of  Oi)eratioii. 

33.  In  Algebra,  the  signs  +,  — ,  and  x  are  used, 
as  in  Arithmetic,  to  represent  addition,  subtraction,  and 
multiplication,  these  operations  being  algebraic,  not 
numerical. 

34.  Signs  of  Addition  and  Subtraction.  The  com- 
bination a-\-h  means  the  algebraic  sum  of  the  quantities 
a  and  5,  and  a  —  h  means  their  algebraic  difference. 


If  «  =  +  4  and  ^>  =  +  3 
If  a  =  +  5  and  d 
If  «  =  —  6  and  ^  =  +  3, 
If  a  =  —  6  and  h 

The  signs  of  addition  and  subtraction  are  the  same  as  those 
used  to  indicate  positive  and  negative  quantities,  but  the  two 
applications  may  be  made  without  confusion,  because  the 
opposite  positive  and  negative  directions  correspond  to  the 
opposite  operations  of  adding  and  subtracting. 

35.  Sign  of  Multiplication.     The  sign  of  multipli- 

I  cation,  x ,  is  generally  omitted  in  Algebra,  and  when 

different  symbols  are  to  be  multiplied,  the  multiplier  is 


EXAMPLES. 

■f  3,     then    a-{-h  = -\-'1,  a—h  = 

+  1. 

—  7,     then     a-\-l  = —'^,  a—h  ■= 

+  12. 

■f  3,     then     a-\-b  =1 —^,  a—h  = 

—  9. 

—  3,     then     «5  +  ^  =  — 9,  «5— Z>  = 

—3. 

14  THE  LANGUAGE   OF  ALGEBRA. 

written  before  the  multiplicand  without  any  sign  be- 
tween them. 

Thus,  4«    means    «  x  4. 

ax        "         X  X  a. 
Sahmy        '*         yxmxbxaxS. 

If  numbers  are  used  instead  of  symbols,  some  sign  of  mul- 
tiplication must  be  inserted  between  them  to  avoid  confusion. 
Thus,  34  would  be  confounded  with  the  number  thirty-four, 
A  simple  dot  is  therefore  inserted  instead  of  the  sign  x . 

Thus,  3.4  =  4x3  =  12. 

3.12-2  =  72. 
1.2.3.4.5  =  120. 
1.2.3.4.5.6  =  720. 
The  only  reason  why  the  point  is  used  instead  of   x ,  is 
that  it  is  more  easily  written  and  takes  up  less  space. 

36.  Division  m  Algebra  is  sometimes  represented 
by  the  symbol  -^,  the  dividend  being  placed  to  the  left 
and  the  divisor  to  the  right  of  this  symbol. 

Ex.     a  -7-b  means  the  quotient  of  a  divided  by  5. 

But  division  is  more  generally  represented  by  writing 
the  dividend  as  the  numerator  and  the  divisor  as  the 
denominator  of  a  fraction. 

Ex.     The  quotient  of  a  divided  by  i  is  written  %• 

It  is  shown  in  Arithmetic  that  a  fraction  is  equal  to  the 
quotient  of  its  numerator  divided  by  its  denominator  ;  hence 
this  expression  for  a  quotient  is  a  vulgar  fraction. 

37.  Powers  and  Exponents,  A  Power  of  a  quan- 
tity is  the  product  obtained  by  taking  that  quantity  a 
certain  number  of  times  as  a  factor. 

Def.  The  Degree  of  the  power  means  the  number 
of  times  the  quantity  is  taken  as  a  factor. 

If  a  quantity  is  to  be  raised  to  a  power,  the  result 
mayf,  in  accordance  with  the  rule  for  multiplication,  be 


SIGI^S   OF  OPERATION.  15 

expressed  by  writing  the  quantity  the  required  number 
of  times. 

Examples.     The  fifth  power  of  a  may  be  written 
axaxaxaxa    or    aaaaa ; 
and  the  fourth  power  of  7,        7.7.7.?  =  2401. 

To  save  repetition,  the  symbol  of  which  the  power  is 
to  be  expressed  is  written  but  once,  and  the  number  of 
times  it  is  taken  as  a  factor  is  written  in  small  figures 
after  and  above  it. 

Thus,  aaaaa    is  written     a^ ; 

7.7.7.7     "        "  74; 

XXX     "  "  x^. 

Dsf.  A  figure  written  to  indicate  a  power  is  called 
an  Exponent. 

Def.    The  operation  of  forming  a  power  is  called /^ 
Involution. 

38.  Hoots.  A  Root  is  one  of  the  equal  factors 
into  which  a  number  can  be  divided. 

Def.  The  figure  or  letter  showing  the  number  of 
equal  factors  into  which  a  quantity  is  to  be  divided  is 
called  the  Index  of  the  root. 

The  square  root  of  a  symbol  is  expressed  by  writing 
the  sign  y"  (called  root)  before  it. 

Ex.  I.     V49  means  the  square  root  of  49,  that  is,  7. 

Ex.  2.     \/x    means  the  square  root  of  x. 

Any  other  root  than  the  square  is  represented  by 
writing  its  index  before  the  sign  of  the  root. 
Ex.  I.     \^x  means  the  cube  root  of  ^. 
Ex.  2.     Vo:  means  the  fourth  root  of  x, 

Def.  The  operation  of  extracting  a  root  is  called 
Evolution. 

39.  The  operations  of  Addition,  Subtraction.  Multi- 
plication, Division,  Involution,  and  Evolution,  are  the 
six  fundamental  operations  of  Algebra. 


16  TEE  ALGEBRAIC   LANGUAGE. 

40.  Bef.  An  Algebraic  Expression  is  any  combi- 
nation of  algebraic  symbols  made  in  accordance  with 
the  foregoing  principles. 

EXERCISES. 

In  the  following  expressions,  suppose  «j  =  —  7,  ^  =  —  5, 
c  zz:  0,  m  =  3,  ^  =  4,  ^-^  =  9,  and  compute  their  numerical 
values. 


I. 

a  +  Z>  +  ^?z  +  7J. 

2. 

a  +  7)1  +  ?Z. 

3. 

m  —  n  —  a  —  h. 

4. 

71  -{-  p  —  m  —  a. 

5- 

3a  —  m  +  ^  —  271. 

6. 

2a  —  7p-^2h  —  m 

7. 

3m  np. 

8. 

7nncp. 

9- 

hm7i. 

10. 

hup. 

II. 

ahnp. 

12. 

2hih7ip. 

13. 

am  +  In. 

14. 

am  —  hi. 

15- 

hp  —  «7Z. 

16. 

6p  -\-  an. 

17. 

n^p  +  wi^Z^.     - 

18. 

7)1^)1  —  ap^. 

19. 

a^  +  R 

20. 

a^  +  h\ 

21. 

a^  -  ¥, 

22. 

ahn  —-  h^n. 

23. 

aW  —  mhi^. 

24. 

aW  -  ¥m\ 

25. 

aW-  +  a%. 

26. 

ah^  —  a%. 

27. 

ah  +  ^^?^ 

28. 

ac  —  hp 
hi  —  7n2) 

20. 

2m%2  _  I0m3 

^0. 

ah  —  mp 

p  —  hc7}i  m  —  n 

In  the  following  expressions,  suppose  oj  =  8,  ^  =  —  3,  and 
X  to  haye  in  succession  the  fifteen  values  — 7,  — 6,  — 5,  etc., 
to  f  7,  and  compute  the  fifteen  corresponding  values  of  each 
expression : 


31.     x^-\-hx-^a.                   32.     

•  hx 

Arrange  the 

results  in  a  table,  thus  : 

x  =  -7; 

Expression  31  =:  78  ; 

Exp.  32  - 

-!!• 

X  —  —6; 

"  =  62  ; 

etc. 

X  —  --6; 

"  —  48. 

etc. 

etc.                etc. 

COMPOUND   EXPRESSIONS.  17 

CHAPTER     III. 
FORMATION    OF   COMPOUND    EXPRESSIONS. 


Fundamental   Principles. 

41.  The  following  are  two  fundamental  principles  of 
the  algebraic  language : 

First  Principle.  Every  algebraic  expression,  how- 
ever complex,  represents  a  quantity,  and  may  be 
operated  upon  as  if  it  were  a  single  symbol  of  that 
quantity. 

Second  Principle.  A  single  symbol  may  be  used 
to  represent  any  algebraic  expression  whatever. 

43.  When  an  expression  is  to  be  operated  upon  as 
a  single  quantity,  it  is  enclosed  between  parentheses, 
but  the  parentheses  may  be  omitted,  when  no  ambiguity 
or  error  will  result  from  the  omission. 

Example.  Let  us  have  to  subtract  h  from  a,  and  multiply 
the  remainder  by  the  factor  m.  The  remainder  will  be  ex- 
pressed hj  a  —  h,  and  if  we  write  the  product  of  this  quantity 
by  m,  in  the  way  of  §  35,  the  result  will  be 

ma  —  h. 
But  this  will  mean  h  subtracted  from  ma,  which  is  not  what 
we  want,  because  it  is  not  a,  but  a  —  b  which  is  to  be  multi- 
plied by  m.     To  express  the  required  operations,  we  enclose 
a  —  h  in  brackets  or  parentheses,  and  write  m  outside,  thus  : 

m  {a  —  b). 

NUMERICAL     EXAMPLES. 

7(8  — 2)  =  7-6  =  42;    but    7-8  —  2  =  56  —  2  in  54 
12(3  +4)  =  12-7  =  84. 
(6  -f3)(2  +  6)  =  9.8  -  72. 
(7  _  4)  (1  -  5)  (2+  7)  =  3  X  -4.9  =  -  108. 
2 


Divided  by  m, 


18  THE  LANGUAGE  OF  ALGEBBA, 

Example  2.  Suppose  that- the  expression  a  —  h-\-c  is  to 
be  added  to  7n,  subtracted  from  m,  multiplied  by  m,  divided 
by  7n,  raised  to  the  third  power,  or  have  the  cube  root  extracted. 
The  results  will  be  written: 

Added  to  7n,  7n  +  {a  —  b  +  c). 

Subtracted  from  m,         m  —  {a  —  b  +  c). 
Multiplied  by  m,  m  (a  —  ^  +  c), 

(a  —  b-i-c) 
m 

Cubed,  {a--b-\-cY. 

Cube  root  extracted,        ^/(a  —  b-\~c). 

There  are  two  of  these  six  cases  in  which  the  parentheses 
are  unnecessary,  although  they  do  no  harm,  namely,  addition 
and  division,  because  in  the  case  of  addition, 

771  -^  {a  —  b  -^  c) 
is  the  same  as  m  -\-  a  —  b  -\-  c. 

[For  example,     10  +  (8  -  5  +  4)  =  10  +  7  ==  17, 
and  10  +  8  —  5  +  4     =  17  also.] 

Again,  in  the  case  of  the  fraction,  it  will  be  seen  that  it  has 
exactly  the  same  meaning  with  or  without  the  parentheses. 

43.  An  algebraic  expression  having  parentheses  as 
a  part  of  it  may  be  itself  enclosed  in  parentheses  with 
other  expressions,  and  this  may  be  repeated  to  any 
extent.  Each  order  of  parentheses  must  then  be  made 
larger  or  thicker,  or  different  in  shape  to  distinguish  it. 

Examples,  i.  Suppose  that  we  have  to  subtract  a  from 
^,  the  remainder  from  c,  that  remainder  from  d,  and  so  ouc 
We  shall  have. 

First  remainder,  h  —  a. 

Second,  c  —  {b  —  a). 

Third,  d-lc-ib-  a)]. 

Fourth,  e-\d---[c-^{b--ay\\. 

Fifth,  f^le-^\d-^[c^{b^a)^\\ 


DEFINITIONS,  19 

2.  Suppose  that  we  have  to  multiply  the  difference  of  the 
quantities  a  and  ^  by  ^  and  subtract  the  product  from  m.  The 
result  or  remainder  will  be 

m  —p{a—  b). 

Suppose  now  that  we  have  to  multiply  this  result  by  p  +  q. 
We  must  enclose  both  factors  m  parentheses,  and  the  result 
will  then  be  written  : 

EXERCISES. 

In  the  following  expressions,  suppose 

a  =z  —1,     b  ^=2  3,    m  =  5.     x  =:  —  3,  —  1,  +  1,  -f  3, 
and  calculate  the  four  values  of  each  expression  which  result 
from  giving  x  the  above  four  values  in  succession. 

x{x  —  a)  {x  —  2a)  {x  —  3a) 

[a{b  —  x)  —  b  {a  —  x)f 
m  (b  —  x)-\-b  {m  —  x) 

3.  \ax  -^  b(x  -^  aY  +  m  (x  —  a)^^ 

4.  [,y^{mx^  _f-  ^)  —  ^(wx^  —  b)]  ^{fnb  —  a). 

Note,     When  the  square  root  is  not  an  integer,  it  will  be  sufficient 
to  express  it  without  computing  it  in  full. 
Thus,  for  X  =  —'dj  we  shall  have 

^{mx^  +  6)  -  ^^{mx^'  -  &)  =  V^S  -  ^^42, 

This  is  a  sufficient  answer  without  extracting  the  roots. 

Defiiiitionso 

44.  Coefficient.  Any  number  which  multiplies  a 
quantity  is  called  a  Coefficient  of  that  quantity.  A 
coefficient  is  therefore  a  multiplier. 

Example.     In  the  expression  4,abx, 

4  is  the  coefficient  of  abx, 
ia    "  "  "  bx, 

4:ab     ''  "  "  X. 


20  THE  LANGUAGE   OF  ALGEBRA. 

Def.  A  Numerical  Coefficient  is  a  simple  ntiinber, 
as  4,  in  the  above  example. 

Def.  A  Literal  CoefEicient  is  one  containing  one 
or  more  letters  used  as  algebraic  symbols. 

Kem.  Any  quantity  may  be  considered  as  having 
the  coefficient  1,  because  \x  is  the  same  as  x. 

Reciprocal,  The  ReciprocaL  of  a  number  is  unity 
divided  by  that  number.     In  the  language  of  Algebra, 

Reciprocal  of  JV  =  -^^ 

Formula,  A  Formula  is  an  expression  used  to 
show  how  a  quantity  is  to  be  expressed  or  calculated. 

Term,  When  an  expression  is  made  up  of  several 
parts  connected  by  the  signs  -h  or  — ,  each  of  these 
parts  is  called  a  Term. 

Example. — In  the  expression, 

a  -^-Ix  -\-  3ma^, 
there  are  three  terms,  a,  bx,  and  dmx^. 

When  several  terms  are  enclosed  between  parentheses,  so 
as  to  be  operated  on  as  a  single  symbol,  they  form  a  single 
term. 

Thus,  the  expression 

{a^  bx  +  Zmx^)  {a  +  h) 

forms  but  a  single  term,  though  both  numerator  and  denom- 
inator are  each  a  product  of  several  terms.  Such  expressions 
may  be  called  compound  terms. 

Aggregate,  A  sum  of  several  terms  enclosed  be- 
tween parentheses  in  order  to  be  operated  upon  as  a 
single  quantity  is  called  an  Aggregate. 

Algebraic  expressions  are  divided  into  monomials 
and  polynomials. 

A  Monomial  consists  of  a  single  term. 


DEFINITIONS.  21 

A  Polynomial  consists  of  more  than  one  term. 

A  Binomial  is  a  polynomial  of  two  terms. 

A  Trinomial  is  a  polynomial  of  three  terms. 

Note.  The  last  three  words  are  commonly  applied  only 
to  sums  of  simple  terms,  formed  of  single  symbols  or  products 
of  single  symbols. 

Entire.  An  Entire  Quantity  is  one  which  is  ex- 
pressed without  any  denominator  or  divisor,  as  2,  3,  4, 
etc.  ;  a,  &5  x^  etc.  ;  2a6,  2mp,  db  {x  —  y\  etc. 

A  Theorem  is  the  statement  of  any  general  truth. 

45.  Other  Algebraic  Signs.  Besides  the  signs  al- 
ready defined,  others  are  of  occasional  use  in  Algebra. 

>,  the  Sign  of  Inequality,  shows  when  placed  be- 
tween two  quantities,  that  the  one  at  the  open  end  of 
the  angle  is  the  greater. 

Ex.  I.     a'yh  means  a  is  greater  than  1). 

Ex.  2.  m  <ix  <^n  means  x  is  greater  than  m,  but  less 
than  n. 

: ,  another  Sign  of  Division,  is  placed  between  two 
quantities  to  express  their  ratio. 

Thus,  ^  :  ^  means  the  ratio  of  a  to  h,  or  the  quotient  of  a 
divided  by  1). 

.*.  means  Hence,  or  Consequently;  as, 

«  +  2  =  5  ;  .-.    a  =  3. 

QO  means  a  quantity  infinitely  great,  or  Infinity. 

,  the  Vinculum,  is  sometimes  placed  over  an 
aggregate  to  include  it  in  one  mass,  in  lieu  of  paren- 
theses. 


Ex.     a—h  c  —  d  is  the  same  as  {a  —  b)  {c  —  d). 

It  is  mostly  used  with  the  radical  sign.    We  often  write 


Va  i-  b  -\-  c    instead  of     V(a  +  b  -\-  c). 


.22  THE  LANGUAGE   OF  ALGEBRA, 

CHAPTER     IV. 

CONSTRUCTION    OF    ALGEBRAIC    EXPRESSIONS. 

46.  All  operations  upon  algebraic  quantities,  however 
complex,  consist  in  combinations  of  the  elementary  operations 
already  described.  The  result  of  each  single  operation  will  be 
an  aggregate,  a  product,  a  quotient,  or  a  root,  and  every  such 
result  may,  in  subsequent  operations,  be  operated  upon  as  a 
single  symbol.  There  are  only  three  cases  in  which  an  expres- 
sion needs  any  modification  in  order  to  be  operated  upon, 
namely : 

Case  I.  An  aggregate  must  be  enclosed  in  parentheses,  if 
any  other  operations  than  addition  or  division  are  to  be  per- 
formed upon  it.     (§  42.) 

Case  II.  When  a  product  is  to  be  raised  to  a  power,  or  to 
have  a  root  extracted,  it  may  be  enclosed  in  parentheses  in 
order  to  show  that  the  operation  extends  to  all  the  factors. 

If  we  take  the  product  ahc,  and  write  an  exponent,  2  for 
instance,  after  it  thus,  ahc^,  it  would  apply  only  to  c,  and 
would  mean  a  x  h  X  c^.  So  with  the  radical  sign  ;  A^ahc 
might  mean  only  yy/a  xhxc.  To  indicate  that  the  power 
or  root  is  that  of  the  product  as  a  whole,  we  may  enclose  it 
in  parentheses,  thus  : 

Square  root  of  abc  =  ^{abc). 
Square  of  abc  =  {abcy. 

But  a  root  sign  is  commonly  made  to  include  the  whole 
product  by  simply  extending  a  vinculum  over  all  the  factors 
of  the  product,  thus :    Square  root  of  abc  =  Vcibc. 

Case  III.  If  negative  quantities  are  to  be  multiplied, 
merely  writing  them  after  each  other  would  lead  to  mistakes. 
Thus,  the  product  ax  —bx—c,  if  written  without  the  x 
sign,  would  be  a  —  b  ^  c,  and  would  not  mean  a  product  at 
all.     But,  by  enclosing  —b  and  —cm  parentheses,  we  have 

a{-b){-c), 
which  would  correctly  express  the  product  required. 


CONSTRUCTION  OF  ALGEBRAIC  EXFRE8SI0N8.      23 

47.  The  following  example  will  show  how  operations  may 
be  combined  to  any  extent. 

The  quantity  a  is  to  be  subtracted  from  t,  and  the  differ- 
ence multiplied  by  y,  forming  a  product  P.*  The  quotient  of 
p  —  r  divided  by  q  is  to  be  multiplied  by  m,  and  the  product 
subtracted  from  P,  The  difference  is  to  form  the  numerator 
iV^  of  a  fraction.  To  form  the  denominator,  h  is  to  be  added 
to  a  and  subtracted  from  it,  and  the  product  Q  of  the  sum  and 
difference  formed.  The  quantity  q  is  to  be  added  to  and  sub- 
tracted from  p,  and  the  product  R  of  the  sum  and  difference 
formed.  The  quotient  of  Q  divided  by  R  is  to  form  the  de- 
nominator of  the  fraction  of  which  the  numerator  is  P. 

The  quantity  I  subtracted  from  a  leaves   h  —  a. 

Multiplying  it  by  y,  the  product  P  is        y  {h  —  a). 

Quotient  of  2^  —  r  divided  by  q  — ■» 

Multiplying  it  by  m,  m 

[If  instead  of  multiplying  the  fraction  as  a  whole  by  7??, 
we  had  multiplied  its  numerator,   we   should  have  had  to 

enclose    the  jt?  —  r   in    parentheses,  thus:    — ^ --•     But 

when  the  multiplier  is  written  at  the  end  of  the  line,  between 
the  terms  of  the  fraction,  as  above,  it  indicates  that  the  frac- 
tion, as  a  whole,  is  multiplied  by  m,~\ 

P  —  T 

Subtracting  the  last  product  from  P,  it  is  y(b —a)—m  — 

Adding  b  to  a,  a  -\-  h 

Subtracting  h  from  a,  a  —  h. 

The  product  Q  of  the  sum  and  difference,  {a  -{-  b)  (a  —  h). 
The  product  R  0^  p  -{-  q  hy  p  —  q,  {P  +  Q)  {P  —  5')- 

The  quotient  of  Q  divided  by  R,  {a-hl?)(a-^ 

(p  +  q){p-  g) 

*  In  mathematical  language,  when  a  substantive  is  followed  by  a 
symbol  in  this  manner,  the  latter  is  used  as  a  sort  of  proper  name  to 
designate  the  substantive,  so  that  the  latter  can  be  afterward  referred  to 
by  the  letter  without  ambiguity. 

In  the  present  case,  the  capital  letters  are  used  in  accordance  with 
the  second  general  principle,  §  41. 


24  THE  LANGUAGE   OF  ALGEBRA, 

The  fraction  having  iV^for  its  numerator  and  this  quotient 

for  its  denominator  is 

(I.        \           p  —  r 
y\b  ^  a)  —  m 

(^  +  ^)  (^  —  h) 

48.  By  the  second  general  principle,  §  41,  a  single  sym- 
bol may  be  written  in  place  of  any  algebraic  expression  whateverc 
When  several  symbols  indicating  such  expressions  are  com- 
bined, the  original  expressions  may  be  substituted  for  them, 
ftnd  be  treated  in  accordance  with  the  first  principle. 

EXAMPLES. 

Suppose       F  z=  a  -{-  bx;  Q  =: ; 

T  =  X  —  y;  V  =  mpq. 

It  is  required  to  form  the  expression 
PQ-TV 

The  answer  is 

(«  +  ^^)  ^^     —  (^  —  y)  m^q 


{a  +  Ix)  (x  —  y)—  ^  ^/^  "^Pq 


EXERCISES. 

Form  the  expressions: 

I.     P—T,  2.  T—P. 

3.     P-Q'  4.  Q-V, 

5.     VP.  6.  V(P  +  n 

7.     y^(p  _  T).  8.  P^T'. 

9.      V\  10.  T^V\ 

VP  ^  QT  PT 

"•        Qi-T^'  '^-  QV' 

(P+  T)(P—T)  {3P  —  2TY 

pa-  y  2(P+  r)^ 

^5-     ^(^p_Tf  '^-  (2T-F)^' 


EXERCISES,  25 


P2r2  PT-\-TQ 

'^*        {P-fTf    '  ^°'     (F-r)(F+2^)' 

F— (g2_  y)2  F-  g^+  y 


EXERCISES    IN    ALGEBRAIC     LANGUAGE. 

The  following  questions  are  proposed  to  practice  the  student  in  ex- 
pressing the  relations  of  quantities  in  algebraic  language.  Should  any 
of  them  offer  difficulties,  he  is  recommended  to  substitute  numbers  for 
the  algebraic  letters,  examine  the  process  by  which  he  proceeds,  and  then 
apply  the  same  process  to  the  letters  that  he  applied  to  the  numbers.  No 
solutions  of  equations  are  required. 

1.  How  many  cents  are  there  in  m  dollars  ? 

2.  How  many  dollars  in  m  cents  ? 

3.  A  man  had  a  dollars  in  one  pocket,  and  h  cents  in  the 
other ;  how  many  cents  had  he  in  all  ?    How  many  dollars  ? 

4.  The  sum  of  the  quantities  a  and  h  is  to  be  multiplied 
by  m.     Express  the  product,  and  its  square. 

5.  A  man  having  I  dollars  paid  out  m  dollars  to  one  per- 
son and  n  dollars  to  another.  Express  what  he  had  left  in 
two  ways  ? 

6.  How  many  chickens  at  h  cents  a  piece  can  be  purchased 
for  m  dollars  ? 

7.  A  man  walked  from  home  a  distance  of  m  miles  at  4 
miles  an  hour,  and  returned  at  the  rate  of  3  miles  an  hour. 
How  long  did  it  take  him  to  go  and  come  ? 

8.  A  man  going  to  market  bought  tomatoes  at  h  cents  per 
peck  and  potatoes  at  k  cents  a  peck,  of  each  an  equal  number. 
They  cost  him  m  cents.     How  many  pecks  of  each  did  he  buy  ? 

9.  How  many  minutes  will  it  require  to  go  a  miles,  at  the 
rate  of  h  miles  an  hour  ? 

10.  A  man  bought  from  his  grocer  a  pounds  of  tea  at  x 
cents  a  pound,  h  pounds  of  sugar  at  y  cents  a  pound,  and^  c 
pounds  of  coffee  at  z  cents  a  pound.  How  many  cents  will 
the  whole  amount  to  ?    How  many  dollars  ?    How  many  mills  ? 

11.  A  man  bought  /  pounds  of  flour  at  m  cents  a  pound. 


26  THE  ALGEBRAIC   LANGUAGE. 

and  handed  the  grocer  an  i?;-dollar  bill  to  be  changed  ?     How 
many  cents  ought  he  to  receive  in  change  ? 

12.  From  two  cities  a  miles  apart  two  men  started  out  at 
the  same  time  to  meet  each  other,  one  going  m  miles  an  hcur 
and  the  other  n  miles  an  hour.  How  long  before  they  will 
meet  ?  How  far  will  the  first  one  have  gone  ?  How  far  will 
the  second  one  have  gone  ? 

13.  A  man  left  his  n  children  a  bonds  worth  x  dollars 
each,  and  b  acres  of  land  worth  y  dollars  an  acre ;  but  he 
owed  m  dollars  to  each  of  q  creditors.  What  was  each  child's 
share  of  the  estate  ? 

14.  Two  numbers,  x  and  y,  are  to  be  added  together,  their 
sum  multiplied  by  s,  that  product  divided*  by  a-\-b,  and  the 
quotient  subtracted  from  h.     Express  the  result. 

15.  The  sum  of  the  numbers  ^j  and  q  is  to  be  divided  by 
the  sum  of  the  numbers  a  and  h,  forming  one  quotient.  The 
difference  of  the  numbers  p  and  q  is  to  be  divided  by  the  dif- 
ference of  the  numbers  a  and  h,  forming  another  quotient. 
The  sum  of  the  two  quotients  is  to  be  multiplied  hj  r-\-s. 
Express  the  product. 

16.  The  quotient  of  x  divided  by  a  is  to  be  subtracted 
from  the  quotient  of  y  divided  by  h,  and  the  remainder  multi- 
plied by  the  sum  of  x  and  y  divided  by  the  difference  between 
X  and  y.     Express  the  result. 

17.  The  number  x  is  to  be  increased  by  6,  the  sum  is  to  be 
multiplied  by  a-\-l),  q  is  to  be  added  to  the  product,  and  the 
sum  is  to  be  divided  by  r  —  s.     Express  the  result. 

18.  A  family  of  brothers  a  in  number  each  had  a  house 
worth  a  thousand  dollars  each.  What  was  the  total  value  of 
all  the  houses  in  dollars  ?     What  was  it  in  cents  ? 

19.  A  grocer  mixed  a  pounds  of  tea  worth  x  cents  a  pound, 
and  h  pounds  worth  y  cents  a  pound.  How  much  a  pound 
was  the  mixture  worth  ? 

20.  x-{-y  houses  each  had  a-\-l)  rooms,  and  each  room 
m-\-n  pieces  of  furniture.  How  many  pieces  of  furniture  were 
there  in  all  ? 

21.  In  a  library  were  j^-f-^'  volumes,  each  volume  hadjf?  +  ^ 
pages,  each  page  p-\-q  words,  and  each  word  on  the.  average 
8  letters.  How  many  letters  were  there  in  all  the  books  of  the 
library  ? 

22.  A  post-boy  started  out  from  a  station,  travelling  Jc 
miles  an  hour.  Three  hours  afterward,  another  one  started 
after  him,  riding  m  miles  an  hour.     How  far  was  the  first  one 


OP  y  EXERCISES.  27 

ahead  oitEe  second  at  the  end  of  x  hours  after  the  second 
started? 

23.  Two  men  started  to  make  the  same  journey  of  in  miles, 
one  going  r  miles  an  hour,  and  the  other  s  miles  an  hour. 
How  much  sooner  will  the  man  going  r  miles  an  hour  make 
his  journey  than  the  one  going  s  miles  an  hour  ?  How  much 
sooner  will  the  one  going  s  miles  an  hour  make  his  journey 
than  the  one  going  r  miles  an  hour  ? 

24.  One  train  runs  from  Boston  to  New  York  in  Ji  hours, 
at  the  rate  of  n  miles  an  hour.  How  long  will  it  take  another 
train  running  5  miles  an  hour  faster  to  perform  the  journey  ? 

25.  If  a  man  bought  h  horses  for  t  dollars,  and  n  yoke  of 
oxen  for  m  dollars,  how  much  more  did  one  horse  cost  than  one 
yoke  of  oxen  ?  How  much  more  did  one  yoke  of  oxen  cost 
than  one  horse  ? 

26.  A  train  making  a  journey  of  2m  miles  goes  the  first 
half  of  the  way  at  the  rate  of  r  miles  an  hour,  and  the  second 
half  at  the  rate  of  s  miles  an  hour.  How  long  did  it  take  it  to 
go  ?    What  was  the  average  speed  for  the  journey  ? 

27.  Two  men,  A  and  B,  started  to  Avalk  from  Hartford  to 
New  Haven  and  back,. the  distance  between  the  two  cities 
being  a  miles.  A  goes  p  miles  an  hoiir  and  B  q  miles  an  hour. 
How  far  will  A  have  got  on  his  return  journey  when  B  reaches 
Hartford  ? 

28.  A  man  having  h  dollars  bought  h  books  at  $6  each. 
How  many  books  at  $4  each  can  he  buy  with  the  balance  of 
his  money  ? 

29.  A  man  going  to  his  grocer  with  m  dollars,  bought  s 
pounds  of  sugar  at  a  cents  a  pound,  and  r  pounds  of  coffee  at 
b  cents  a  pound.  How  many  barrels  of  flour  at  q  dollars  a 
barrel  can  he  buy  with  the  balance  of  his  money  ? 

30.  A  man  divided  m  dollars  equally  among  a  poor  Chinese 
and  n  dollars  equally  among  h  orphans.  Two  of  the  Chinese 
and  three  of  the  orphans  put  their  shares  together  and  bought 
X  Bibles  for  the  heathen.     How  much  did  each  Bible  cost  ? 

31.  A  pedestrian  having  agreed  to  walk  the  a  miles  from 
Boston  to  Natick  in  h  hours,  travels  the  first  k  hours  at  the 
rate  of  m  miles  an  houi*.  At  what  rate  must  he  travel  the 
remainder  of  the  time  ? 

32.  A  train  having  to  make  a  journey  of  x  miles  in  li  hours, 
ran  for  k  hours  at  the  rate  of  r  miles  an  hour,  and  then  made 
a  stop  of  m  minutes.  How  fast  must  it  go  during  the  remain- 
der of  its  journey  to  arrive  on  time  ? 


BOOK    II. 
ALGEBRAIC     OPERA  TIONS. 


Creneral   Remarks. 


The  algebraic  expressions  formed  in  accordance  with  the 
rules  of  the  preceding  book  admit  of  being  transformed  and 
simplified  in  a  variety  of  ways.  This  transformation  is  effected 
by  operations  which  have  some  resemblance  to  the  arithmetical 
operations  of  addition,  subtraction,  multiplication,  and  division, 
and  which  are  therefore  called  by  the  same  names. 

In  performing  these  algebraic  operations,  the  student  is  not, 
as  in  Arithmetic,  seeking  for  a  result  which  can  be  written  in 
only  one  way,  but  is  selecting  out  of  a  great  variety  of  forms  of 
expression  some  one  form  which  is  the  simplest  or  the  best  for 
certain  purposes.  Sometimes  one  form  and  sometimes  another 
is  the  best  for  a  particular  problem.  Hence,  it  is  essential 
that  the  algebraist,  in  studying  an  expression,  should  be  able  to 
see  the  different  ways  in  which  it  may  be  written. 

Definitions. 

49.  Function,  An  algebraic  expression  containing 
any  symbol  is  called  a  Function  of  the  quantity  repre- 
sented by  that  symbol. 

Ex.  I.  The  expression  Zx^  is  a  function  of  x, 

Ct  -\-  X 

2.  The  expression is  a  function  of  x  and  also  a 

function  of  a. 

When  an  expression  contains  several  symbols,  we  may 
select  one  of  them  for  special  consideration,  and  call  the  ex- 
pression a  function  of  that  particular  one.  For  instance, 
although  the  expressions. 


DEFINITIONS.  39 

a  ■\-hx^  -\-  cx% 

tn  +  nV'^', 

contain  other  symbols  besides  x,   they  are  both   functions 

of  X, 

50.  An  Entire  Function  is  one  in  which  the  quan- 
tity is  used  only  in  the  operations  of  addition^  subtrac- 
tion and  multiplication. 

Example.     The  expressions 

ax  +  y, 

(«2  ^  ^2)  ^  _  (Jji  ^  y'jx^—X  +  d, 

are  entire  functions  of  x.     But  the  expressions 

^.^+1    and    3V~x 
ax  —  y 

are  not  entire  functions  of  x,  because  in  the  one  x  appears  as 
part  of  a  divisor,  and  in  the  other  its  square  root  is  extracted. 
An  entire  function  of  x  can  always  be  expressed  as  a  sum 
of  terms,  arranged  according  to  the  powers  of  x  which  they 
contain  as  factors.     The  form  of  the  expression  will  then  be 

A  +  BX+  Cx^  j^  Dx^  +  Ex^-^  etc., 
where  A,  B,  (7,  etc.,  may  represent  any  algebraic  expressions 
which  do  not  contain  x. 

51.  Like  Terms  are  those  which  are  formed  of  the 
same  algebraic  symbols,  combined  in  the  same  way, 
and  differ  only  in  their  numerical  coefficients. 

Ex.     The  terms  ax,  2ax,  —6ax  are  like  terms. 

52.  The  Degree  of  any  term  is  the  number  of  its 
literal  factors. 

Examples.  The  expression  ahxy  is  of  the  fourth  degree, 
because  it  contains  four  literal  factors. 

The  expression  x^  is  of  the  third  degree,  because  the  letter 
X  is  taken  three  times  as  a  factor. 

The  expression  ah^x^  is  of  the  sixth  degree,  because  it  con- 
tains  a  once,  h  twice,  and  x  three  times  as  a  factor. 

When  an  expression  consists  of  several  terms,  its 
degree  is  that  of  its  highest  term. 


30  ALGEBRAIC   OPERATIONS. 

CHAPTER     I. 

ALGEBRAIC    ADDITION    AND    SUBTRACTION. 


Algebraic  Additiou. 

53.  By  the  language  of  Algebra,  the  sum  of  any  number 
of  quantities,  positive  or  negative,  may  be  expressed  by  writing 
them  in  a  row,  with  the  sign  +  before  all  the  positive  quan- 
tities, and  the  sign  —  before  the  negative  ones. 

Ex.  A-^B—D—X-\-  Y,  etc.,  is  the  algebraic  sum  of  the 
several  quantities  A,  B,  —D,  —X,  Y,  etc. 

54.  To  simplify  an  expression  of  the  sum  of  several 
quantities, 

1.  AVhen  dissimilar  terms  are  to  be  added,  no  sim- 
plification can  be  effected. 

Ex.  If  we  require  the  sum  of  the  five  expressions,  «,  — xy^ 
mp,  nq,  and  —hits,  we  can  only  write, 

a  —  xy  -\-  mp  -\-  nq  —  Ihs, 
according  to  the  language  of  Algebra,  and  cannot  reduce  the 
expression  to  a  simpler  form. 

2.  If  mere  numbers  are  among  the  quantities  to  be 
added,  their  algebraic  sum  may  be  formed. 

Ex.  The  sum  of  the  five  quantities  —8,  ah,  5,  mnp,  —15, 
is  found  to  be  —  18  +  ah  -f  wnp, 

3.  When  several  terms  are  similar,  add  the  coeffi- 
cients and  affix  the  common  symbol  to  the  sum. 

When  no  numerical  coefficient  is  written,  the  coefficient 
4-1  or  —1  is  understood.     (§  44.) 

EXAMPLES. 

«  4-  a  r=  2^  [because  1  +  1  =  2]. 
2a  —  a  :=z  a  [because  2  —  1  ==  1]. 


ALGEBRAIC   ADDITION.  31 

3a  +  4(^  —  7«  =  0  [because  3  +  4  —  7  r::z  0]. 

a  -{■  2x  —  3a  —  6x=z  — 2a — 3x  [adding  the  a's  and  the  x^s], 

—  3axy  +  ibm  —  2axy  -}-  bm  =  —  baxy  -f-  bhm. 

Add  the  expressions, 

1.  Ix  +  bhy\  2x  —  3by^  -- 4:X  -  bbyS  6x  —  by%  x  —  %l 

WORK. 

For  convenience,  the  several  terms  may  be  ry^    i    Kj.y2 

written  under  each  other,  as  in  the  margin.     The  g/r '^hiJ^ 

coefficients  of  x  are  7,  3,  —4,  5,  and  1,  of  which  a     ^jT^.i 

the  algebraic  sum  is  11.     The  coefficients  of  y'^  p-     7^2 

are  5,  —3,  —5,  —1,  —1  ;  the  sum  is  —5.     Hence  A 

the  result.  ^- 

Stt7n,  11:?;  —  bby^ 

2.  Sax  —  y  —  2y  +  6,  7ax—y  —  d-\-am,  2ax—y—3-\-6p, 
Here  2x,  am,  and  p,  work. 

all  being  different  sym-  g^^2 y '2x  4-  b 

bols,  the  terms  contain-  i*/^^2 ^  g    ,    ^^^ 

ing  them  do  not  admit  _    ^^2  —    y             —  3               -1-5;^ 

of  simplification  (§  54,  ^ 

1).      The    numbers    5,  Smn,    —  3y  —  2x  —  7  -f  am  +  bp 

—9.  —3,  are  added  by 

the  rule  (§  54, 2).     The  coefficients  of  ax^  cancel  each  other  (8—7—1  =  0). 

3.  Add  6{x-{~  y\  5  {x  +  ^)  +  a,  2  {x  H-  y)  -  3a, 

Here  the  aggregate,  x-\-y,  enclosed  in 
parentheses,  is  treated  as  a  simple  symbol.  ,  ' 

Note.     When   the   student   can  add  b  (x  ■\-  y)  ^ 

the  coefficients  mentally,  it  is  not  neces-  ^  "^  .^ 

sary  to  write  the  expressions  under  each "^ 

other.     Nor  is  it  necessary  to  repeat  the       Sum   13  (o;  +  ^)  2a 

symbol  after  each  coefficient. 

EXERCISES. 

1.  3a  +  75  —  8^  +d,3a  —  2bfc  —  e,  ^a  —  b  —  c—d, 

2.  7a  — {x  -\-  y),  Sa  —  {x  +  y)^  3  {x  -f  y)  —  16^. 

3.  7oy^  —  2x  —  5,  2x^  —  3:r  -j-  8,   —  9.^2  H-  bx  +  3. 

4.  x'^  •+  2x  —  y,  ix^  +  7.1'  —  2y,  —  2:1'^  +  :r  —  9^/,    —  3x'- 
^x  -—  y. 

5.  9  (a-  +  by,  10  (a  +  b)%  {a  +  b)%  2  {a  +  b)%  ^x-y-z. 

6.  2(m  +  n)  +  3  (aj  +  ^>),     («  +  &)  —  (m  +  n),     (a  +  b) 
—  (m  -f  n). 

7.  7^3  —  2a2  ^  3^^^   _  ^3  _  ^2  _  ^^^   _  6^3  j_  3^2  _  2«i^;. 


32  ALGEBRAIC   OPERATIONS. 

8.  (m  +  nf  +  X,  2  (m  +  7iY  —  y,  3  {7n  +  nf  —  2a;, 
(m  +  7i)^  -—  y- 

9.  (i^  +  ^Y  -  6>  (i^  +  ^)'  +  ^^  (^  +  ^)'  +  ^.  (i^  +  ^)H^. 

10.  6a{x  —  y),  ba(x  —  y),  2a{x  —  y),  a{x  —  y), 

11.  2  (m  —  7^)  a;  4-  2,  3  (m  +  t^)  a;  —  5,  b{m  +  n)x  —  Q, 
7  {m  -{-  n)x  —  8. 

^-    .,    ^^  9^A.^y  ^    y  y    ^  ^    ^ 

a       a         ha       h'  h       7    a       7 

ir      m    ^x      ^m     ^x      ^m     .x       ^  m 

11. ,  2 2  —  ,  3 3  —  .  4 4  — 

y      n       y         n        y         n        y         n 

m  -\-  n        m  +  n        m  +  n         m  +  n 

15.  Of  two  farmers,  the  first  had  2x  —  dy  acres,  and  the 
second  had  x  —  y  acres  more  than  the  first.  How  many  acres 
had  they  both? 

16.  A  had  2x  dollars,  B  had  y  dollars  less  than  A,  and  C 
had  2y  dollars  more  than  A  and  B  together.  How  many  had 
they  all  ? 

17.  A  father  gave  his  eldest  son  x  dollars,  his  second  5  dol- 
lars less  than  the  first,  his  third  5  dollars  less  than  his  second, 
and  his  fourth  5  dollars  less  than  his  third.  How  much  did 
he  give  them  all  ? 

55.  Addition  with  Literal  Coefficients,  When  dif- 
ferent terms  contain  the  same  symbol,  multiplied  by 
diiferent  literal  coefficients,  these  coeflScients  may  be 
added  and  the  common  symbol  be  aflBixed  to  their 
aggregate. 

EXAM  PLE  S. 

1.  As  we  reduce  the  polynomial 

'  6x  -{-  6X'—2x 

to  the  single  term      (6  +  5  —  2)  ^  =  3a^, 
so  we  may  reduce  the  polynomial 

ax  ■\-hx  -^  ex 
to  the  single  term,  (a  -^-h  —  c)x. 

2.  The  expression 

mx  -i-  ny  —  bx  -{-  dy  -\-  a  +  b 
may  be  expressed  in  the  form 

{m  —b)  X  -{-  {?i  4-  d)  y  +  a  -\-h 


SUBTRACTION.  33 

EXERCISES. 

Collect  the  coefficients  of  x  and  y  in  the  following  ex- 
pressions : 

1.  ax  -{-ly  -\-  mx  +  ny, 

2.  mnx  +  %by  +  lyqx  —  Aiy, 

3.  dx  —  2y-\-  6bx  -—  4^  +  7ax  +  m  -{-  n, 

4.  Sax  -f  8^.-?;  -}-  by  -{-  7x  —  6y  -{-  x  —  by, 

5.  a:r  +  %  ■+■  c;2  —  r/7iC  —  ny  —  pz. 

6.  2dx  +  3ey  +  4/2;  —  2fx  —  3<i^  +  4e;2;. 

2  3 

7.  ~«y  — 2:i;  + j%  +  6ax. 

8.  2aa;  —  by  —  dbx  —  4«r?/. 

2  1 

10.  4/wa;  +  2^/  —  3«a;  —  6cx  -{-  ay  —  -mx  +  -dx, 

1 1.  6abx  —  3m/^^/  —  a^;r  +  4:cdy  —  dx. 

12.  Say  +  2bx  —  jdx  +  2ay  —  ^bx, 

1  3 

13-     ^(^y  -  3a:  +  2y  —  -«y  —  52;  +  y. 

14.  3wa.  —  «a:  —  -^y  +  a;  +  c/:c  —  y, 

15.  3aZ>cr  —  my  +  2(:a/^  —  dy  -\-  Vx. 

16.  hm^y  —  6a;  +  4a/^  —  3A/a;  —  y  +  Vy. 

1 7.  4^/0;  —  6«/  +  aVy  +  ex  —  Vy  —  4a  V^/  +  Va:. 

Algebraic  Subtraction. 

56.  Def.  Algebraic  Subtraction  consists  in  ex- 
pressing the  difference  of  two  algebraic  quantities. 

jRule  of  Subtraction.  It  has  been  shown  (§  21)  th^t 
to  subtract  a  positive  quantity,  &,  is  the  same  as  to 
add,  algebraically,  the  negative  quantity,  —6.  Also, 
that  to  subtract  —  &  is  equivalent  to  adding  +6.  Hence 
the  rule : 

Change  the  algebraic  sign  of  all  the  terms  of  the 
subtrahend,  or  conceive  them  to  be  changed,  and  then 
proceed  as  in  addition. 


34  .  ALGEBRAIC   OPERATIONS. 


NUMERICAL      EXAMPLES. 

Min.,    10  +  6  =  16     10+   6  =  16     10+   6  =  16     10+   6  =    16 

Subt.,  _9 ^=_9       9—  4=  5       9—  8=  1       9  — 12  =  -3 

Eem.,     1  +  6=  7      1  +  10=11      1  +  14=15      1  +  18=  I9 

ALGEBRAIC      EXERCISES. 


I.  From  ^x  —  4:ay  +  5^  +  c, 

Subtract  x  —  7a2/  —  8b  -i-  d. 


WORK. 


Minuend,  3x  —  4:ay  +    5^  +  c 

Subtrahend  with  signs  changed,     —  x  -\-  7ay  +    Sb  —  d 

Difference,  2x  +  Say  +  13b  +  c  —  d 

Next  we  may  simply  imagine  the  signs  changed. 

2.  From        7x  —    Uxy  —  12cy  +    8&  +  3ac 
Take         2x  +    Ibxy  +    8cy  —    6b  —  2d 

Diff.,   6x  —  11^2:^  —  20cy  +  Idb  +  3ac  +  2d 

3.  From        8«  +  9<^  —  12c  —  ISd  —  4:X  +  3cy 
Take       I9a  —  7^  —    Sc  —  2hd  H-  3a:  —  A:y 

4.  From         257;?  +  20W  +  92?/  +  35aa;  —    6 
Take         140^  —    82^^  +  20z/  +  %2ax  +  14 

5.  From  8a  +  14*^  subtract  6«  +  20^. 

6.  From  a  —  J  +  c  —  J  take  — a-\-b  —  G-^d. 

7.  From  8«  —  2^  +  3c  subtract  4:a  —  Q>b  —  c  —  2d, 

8.  From  2x^  —  8a;  —  1  subtract  bx^  —  6.t  +  3. 

9.  From  42;4  ^  3:z;3  _  2^2  —  7a;  +  9  subtract  a;^— 2a;3--2a> 
+  7a?  -  9. 

^10.  From  2x^  —  2aa;  +  3a^  subtract  x^  —  ax  +  ri^l 

11.  From  «3  _  3^2^  ^  3^^2  _  ^3  subtract  —  «3  ^  3^2^. 

12.  From  7a:8  _  2:?;2  +  2a;  +  2  subtract  4a;3..,2a.^— 2a;— 14. 

13.  From  b{x  —  y)  -\-  1  {x  —  z)  +  9  (;>;— a;)  take    9  (a;  —  y) 
+  7  (a;  —  2;)  +  5  (2;  -  a;). 

14.  From  l2{a-'b)  —  3(a^h)-{-'ia—  2b  take  7  {a—b) 
--  5  (a  +  ^>). 

15.  From  7^-11^-15^  take  -- 5- +  6  ^  -  7  -  +  8  ^ 

y  z  X  y         z  X         b 


SUBTRACTION,  35 

Clearing  of  Parentheses. 

57.  In  §  42,  2y  it  was  shown  that  an  aggregate  of  terms  in- 
cluded between  i)arentheses  might  be  added  or  subtracted  by 
simply  writing  +  or  —  before  the  parentheses. 

When  an  aggregate  not  multiplied  by  a  factor  is  to  be  added 
or  subtracted,  the  parentheses  may  be  removed  by  the  rules 
for  addition  and  subtraction,  as  follows: 

58.  Plus  Sign  he/ore  Parentheses.  If  the  paren- 
theses are  preceded  by  the  sign  +,  they  may  be 
removed,  and  all  the  terms  added  without  change. 

Example  i.    27+(8-~5— 4  +  7)  =  27  +  8— 5-4  +  7  =  33. 

2.  m-{-{a  —  X— y  +  z)=i7n-\-a^x  —  y-^z, 

3.  22;  +  (—  3a;  —  by)  +  m—^a)  +  (2y— 26?) 

—  ^x  —  ^x  —  by-^^y  —  4:a-\-2y  —  2a 

—  —  X  —  6a. 

The  sign  +  which  precedes  the  parentheses  should  also  be 
considered  as  removed,  but  if  the  first  term  within  the  paren- 
thesis has  no  sign,  the  sign  +  is  understood,  and  must  be 
written  after  removing  the  parentheses, 

EXERCISES. 

Clear  of  parentheses  and  simplify 

1.  x  —  y+  (x  +  y). 

2.  X  -i-  y  -i-  {y  —  x). 

3.  3ab  —  %mp  +  {ah  —  ox  —  2mp). 

4.  2ax  —  3by  +  {7nx  —  2ax  —  pz  -\-  Sby). 

'•    'f  +  (|-^3  +  (5  +  ^^)- 

59.  Minus  Sign  before  Parentheses.  If  the  paren- 
theses are  preceded  by  the  sign  — ,  they  may  be 
removed  and  the  algebraic  sign  of  each  of  the  included 
terms  changed,  according  to  the  rule  for  subtraction  in 

§  56. 

EXAMPLES. 

I.  27-(8-5-4  +  7)=:27  —  8  +  5  +  4  —  7  =  21; 
that  is,  27  —  6  =  21. 


36  ALGEBRAIC   OPERATIONS, 

2.  m  —  {—  a— p-\-y  +  x)  =:m-\-a-\-p  —  y--x. 

3.  3«  4-  X  —  (2a  —  bx)  —  {9x  —  a)  =1  3a  -{-  x  —  2a  +  6x 
—  9x  -{-  a. 

Simplifying  as  in  §  54,  this  reduces  to  2a  —  Sx, 

EXERCISES. 

Clear  the  following  expressions  of  parentheses  and  reduce 
the  results  to  the  simplest  form  by  the  method  of  §  54. 

1.  ab  —  (m  —  Sab  +  2ax)  —  7a^. 

2.  X  —  {a  —  x)  -{-  {x  —  a). 

3.  2b  -i  {b  —  2c)  —  {b+  2c). 

4.  ix  —  3i/  -^  2z  —  {—  ^x  -{■  6y  —  dz)  —  (x  —  y). 

5.  lax  —  2by  —  {^ax  -f  3by)  —  {^ax  —  3by), 

6.  {a  —  x)  ^  {a  -^  x)  -\-  2x. 

7.  —  {a  —  b)  —  (b  —  c)  —  {c  —  a). 

8.  —  (3m  +  ^n)  —  {3m  —  2n)  -f  9m. 

60.  We  may  reverse  the  process  of  clearing  of  parentheses 
by  collecting  several  terms  into  a  single  aggregate,  and  chang- 
ing their  signs  when  we  wish  the  parentheses  to  be  preceded 
by  the  minus  sign.  The  proof  of  the  operation  is  to  clear  the 
parentheses  introduced,  and  thus  obtain  the  original  expression. 

EXERCISES. 

Eeduce  the  following  expressions  to  the  form 
X  —  {an  aggregate). 

1.  X  —  a  —  b.  Ans.  x  —  {a  -\-  b), 

2.  X  —  m  —  n. 

3.  a  -\-  X  —  3x  -{-  2y.  Ans.  x  —  { —  a  -\-  3x  —  2y), 

4.  —3b-Yx-\-2c^  6d. 

5.  2x--2a  -i-  2b.  Ans,  x  —  {— x  -\- 2a  —  2b). 

6.  2x  -^  a  —  b. 

7.  3x  —  2m  +  2n. 

8.  3x  -\-  ab  —  m  —  3ab  +  2m. 

9.  X  —  2m  —  {3a  —  2b).       Ans.  x  —  {2m  -{-  3a  —  2b)o 

10.  X  -{-  3  —  {a  -}-  b). 

11.  X  -}-  a  —  {b  —  c)  -{■  {m  —  n). 

12.  X  —  {am  +  b)  —  {])  —  q)  —  {am  —  n). 

13.  x—  {a  -\-  b)  —  {p  —  q)  —  {m  —  n). 


SUBTMACTIOJV,  *  37 

Coini>ound  Parentheses. 

61.  When  parentheses  of  addition  or  subtraction  are  en- 
closed between  others,  they  may  be  separately  removed  by  the 
preceding  rules. 

We  may  either  begin  with  the  outer  ones  and  go  inward^ 
or  begin  with  the  inner  ones  and  go  outward. 

It  is  common  to  begin  with  the  inner  ones. 

EXAMPLES. 

Clear  of  parentheses: 

1.  f-[e-\d-[c-ib-a)]\l 

Beginning  with  the  inner  parentheses,  the  expression  takes, 
in  succession,  the  following  forms: 

f-[e-  \d-  [c-b  +  a]}'] 
=  f—le—  {d  —  c  +  b  —  afj 
=  f-le-d+  c-b  i-a^ 
=  f  —  e-i-d  —  c  +  b  —  a, 

2.  x^[—  {a  -^  b)  +•  {m  +  ^)  —  {^  —  y)]. 

Eemoving  the  inner  parentheses,  one  by  one,  we  have, 

X  —  [—  a  —  b-\-m-]-n  —  x-\-y\ 
^x-{-a-{-b— m  —  7i-{-x  —  y. 

EXERCISES. 

Kemove  the  parentheses  in  the  following  expressions,  and 
combine  terms  containing  x  and  y,  as  in  §§  54  and  55. 

1.  m  +  [-(p-q)  ^  (a-b)  +  {-c  +  d)l 

2.  m  —  [—  {a  —  b)  —  {p  +  q)  +  {n  —  Jc)]. 

3.  7  ax  —  [{2ax  +  by)  —  {3ax  —  by)  +  {— 7ax  +  2by)]. 

4.  a  —  [a  —  \a  —  [a  —  {a  —  a)]\^. 

5.  j>j  —  [a  —  ^>  —  (5  +  /  +  ^)  +  (—  m  —  n)]. 

6.  2ax  —  [Sax  —  by  —  {7ax  +  2by)  —  {6ax  —  3by)], 

7.  ax-{-by-\-  cz  +  [2ax—3cz  —  {2cz-}-5ax)  —  C^by—Scz)], 

8.  x-^  \  2x  —  y  —  [3x  —  2y  —  {4x  —  3y)]  \. 

9.  ax  —  bz  —  \  ax  -{-  bz  —  [ax  —  bz  —  (ax  +  bz)] }. 
10.     7ny  —  {x  -^  3y  -\-  [2my  —  3  {x  —  y)  —  iab']  +  5 f. 


38  ALGEBBAIG   OPERATIONS, 


II 

12 

13 
14. 

IS 


ax  +  4icx  —  (mic  +  c^  —  2/)  +  [^^^  —  (px  4-  2/)]. 
3^a^  _  Ux  —  (—  3a«/  —  'daz  +  3%)  —  Uz, 
Idax  +  2:^y  —  d  —  [7ad  f  (a;?/  -f  t?)]  —  4:xy. 
m  -\-  4:X  —  [—  4?/  H-  2^  4-  (^^  —  ^)  +  p]- 
2aVy  —  3m  —  [^Va;  —  6^  +  {Vy  —  2a/^)]. 


CHAPTER    II. 

MULTI  PLICATION. 

63.  The  product  of  several  factors  can  always  be 
expressed  Iby  writing  them  after  each  other,  and  enclos- 
ing those  which  are  aggregates  within  parentheses. 

E  X  A   \  T  p  L  E  S . 

The  product  ot  a  +  b  hj  c  =^  c  {a  +  b). 
The  product  of  — ^r-^  by  x  —  y  =  {x  —  y)  — ^• 
The  product  of  a  \-  l  hj  c  -\-  d  z=  i^c  -\-  d)  (a  -\-  b). 
Such  products  may  be  transformed  and  simphfied  by  the 
operation  of  algebraic  multiplication. 

General  Laws  of  Multiplication. 

63.  Law  of  Commutation,  Multiplier  and  multi- 
plicand may  be  interchanged  without  altering  the 
product. 

This  law  is  proved  for  whole  numbers  in  the  following  way. 
Form  several  rows  of  quantities,  each  represented  by  the 
letter  a,  with  an  equal  number  in  each  row,  thus, 

a        a        a        a        a        a 

a        a        a        a        a        a 

a        a        a        a        a        a 

a        a        a        a        a        a 

a        a        a        a        a        a 


MULTIPLICATION.  39 

Let  m  be  the  number  of  rows,  and  n  the  number  of  a's  in 
each  row.     Then,  counting  by  rows  there  will  be 
m  X  n  quantities. 
Counting  by  columns,  there  will  be 
n  X  m  quantities. 
Therefore,  m  x  n  =  n  x  m, 

or  nm  =  mn, 

64.  Law  of  Association,  When  there  are  three 
factors,  m,  n,  and  a, 

771  {no)  =  {7nn)a. 

Example.        3  x  (5  x  8)  =  3  x  40  =  120. 
{3x5)x8  =  15  +  8  =  120. 

Proof  for  Wliole  Numbers.  If  a  in  the  above  scheme 
represents  a  number,  the  sum  of  each  row  will  be  na.  Because 
there  are  m  rows,  the  whole  sum  will  be  7n  (no). 

But  the  whole  number  of  «'s  is  inn.    Therefore, 
m  (na)  =  {mn)  a. 

65.  The  Distributive  Law.  The  product  of  an  ag- 
gregate by  a  factor  is  equal  to  the  sum  of  the  products 
of  each  of  the  parts  which  form  the  aggregate,  by  the 
same  factor.     That  is, 

711  {jo  -\-  q  -{-  r)  =  mj)  -\-  jnq  +  mr.  (1) 

Proof  for  Whole  Numhers.  Let  us  write  each  of  the  quan- 
tities p,  q,  r,  etc.,  m  times  in  a  horizontal  line,  thus, 

p  ^  p  -\-  p  -^  etc.,  m  times  =  mj)* 

q  -^  q  -{-  q  -\-  etc.,  m.  times  =  mq. 

r  +  r  +  r  +  etc.,  7n  times  =  7nr, 

etc.  etc.  etc. 

If  we  add  up  each  vertical  column  on  the  left-hand  side, 
the  sum  of  each  will  be  ^  +  §'  +  r  +  etc.,  the  columns  being 
all  ahke. 

Therefore  the  sum  of  the  m  columns,  or  of  all  the  quantt 
ties,  will  be  , 

m{p  -\-  q  -\-  r,  etc.). 


40  ALGEBRAIC   OPERATIONS. 

The  first  horizontal  line  of  ^'s  being  mp,  the  second  mq, 
etc.,  the  sum  of  the  right-hand  column  will  be 

7np  +  mq  '\-  mr,  etc* 

Since  these  two  expressions  are  the  sums  of  the  same  quan- 
tities, they  are  equal,  as  asserted  in  the  equation  (1). 

Multiplication  of  Positive  Monomials. 

66.  Rule  of  Exponents.     Let  us  form  the  product 

X''^  X  x\ 

By  §  37,     x""  means  xxx,  etc.,  taken  m  times  as  factor. 
X"  means  xxx,  etc.,  taken  7i  times  as  factor. 

The  product  is  xxxxx,  etc.,  taken  {7n-\-n)  times  as  factor. 
Therefore,  x"^  x  x''  =  x'''+\ 

Hence, 

Theorem.  The  exponent  of  the  product  of  like  sym- 
bols is  the  sum  of  the  exponents  of  the  factors. 

67.  As  a  result  of  the  laws  of  commutation  and 
association,  the  factors  of  a  product  may  be  arranged 
and  multiplied  in  such  order  as  will  give  the  product 
the  simplest  form. 

68.  Any  product  of  monomials  may  be  formed  by 
combining  these  principles. 

Example.    Multiply  bniii^x^y^  by  "Ihnx^y, 

By  the  rules  of  algebraic  language,  the  product  may  be  put 
into  the  form 

6mn^x^y^7  bn  x^ y. 

By  interchanging  the  factors  so  as  to  bring  identical  sym- 
bols together, 

b'Umn^noi^x^y^y.  | 

Multiplying  the  numerical  factors  and  adding  the  exponents,, 
the  product  becomes 

35  hmn^oi^  y^. 


\ 


MULTIPLICATION.  41 

69.  We  thus  derive  the  following 

Rule.  Multiply  the  numerical  coefficients  of  the 
factors,  affix  all  the  literal  parts  of  the  factors,  and  give 
to  each  the  sum  of  its  exponents  in  the  separate  factors. 

E  X  E  RC  I  s  E  s.. 

1.  Multiply  xy  by  oi^y.  A7is,  'X^y\ 

2.  Multiply  ^ax  by  2ahx^.  3.  Multiply  hm^y  hy^m^x, 
4.  Multiply  21m?/ by  2«2m.  5.  Multiply  2«m  by  2ma. 
6.  Multiply  5x'^yh  by  xh/h.  7.  Multiply  dxyz  by  dxyz. 

8.  Multiply  2abm  by  2mba.  9.  Multiply  Sab^x^  by  Samv. 

10.  Multiply  2'6mpqr  by  2'6pqrs, 

11.  Multiply  12axy  by  12xyz, 

3  2  3 

12.  Multiply -m^^  by -m^/,       13.  Multiply -r /^^^  by  4m^. 

7 
14.  M.u\tiip\j  -abcdhj  Mefy. 

70.  When  we  have  to  find  the  product  of  three  or  more 
quantities,  we  multiply  two  of  them,  then  that  product  by  the 
third,  that  product  again  by  the  fourth,  and  so  on. 

Ex.  2db  X  2a^'b  x  '^aW'  x  ^bmxy  =  36a^b^mxy, 
Exercises.     Multiply 

1 5 .  mx  X  my  x  mz,  16.     axxbxxcxx  dx, 

1 7.  do?m  X  4:Pn  X  m?i,  18.     abx  2bc  x  ^ca, 

19.  %myi^  X  hwp^  x  9;)m^ 

20.  abxacxadx am3  xyx 2yz x zx, 

21.  amx  X  anx  x  amxy  x  anxy  x  amxyz, 

22.  a^x  X  a^y  x  ax^  x  ay'^  x  «V  x  a^y^  x  x^y^, 

23.  2am  X  San  xa^x  m^  x  ^mx  x  2nx, 

Rule  of  Sig:ns  in  Multiplication. 

71.  It  was  shown  in  §  25  that  a  product  of  two  factors  is 
positive  when  the  factors  have  like  signs,  and  negative  when 
they  have  unlike  signs.     Hence  the  rule  of  signs, 

+    X  +  makes  +? 

+    X  -         "  -, 

-   X  +         "  -, 

~   X  —        "  +. 


42 


ALGEBRAIC   OPERATIONS, 


Examples.     The  quantity  a 

Multiplied  by      3     makes     +  3a. 

2         ''  +2a. 


u 

iC 

1 

u 

+    a. 

6i 

u 

0 

Ci 

0. 

Ci 

a 

-1 

66 

—    a. 

U 

a 

-2 

U 

—  2a. 

The  quantity  —  a 

Multiplied  by 

3 

makes 

—  3a. 

u 

u 

2 

a 

—  2a. 

u 

u 

1 

u 

—    a. 

(6 

ii 

0 

a 

0. 

iC 

u 

-1 

u 

+    a. 

u 

u 

-2 

6C 

+  2a. 

72.  Geometrical  Illustration  of  the  Rule  of  Signs,    Suppose 

the  quantity  a  to  represent  a  length  of  one  centimetre  from 

the  zero  point  toward  the  right  on  the  scale  of  §  11. 

Then  we  shall  have 

0 

a  =  this  line    \  r 


The  product  of  the  line  by  the  factors  from  -[-3  to  —3 
will  be 

0 

a  X  3,  p 


a  X  2, 
a  X  1, 
a  X  0, 
a  X  —  1, 
a  X  —  2y 
a  X  —  3,      ■ 


1 


T 


I 


0 

0 

0 


We  shall  also  have 


a  z=z  this  line 


MULTIPLICATION.  43 


The  products  by  the  same  factors  will  be 

0 

—  a  x^,      I  I  I  I 

0 

—  ax%,  I  I  I 

0 

—  ri?  X  0,  0 

0 

—  ax—1,  I  I 

0 

—  ax  —  2,  I  I  I 

0 

—  a  X  —  3,  f 


I  I  I  i 

These  results  are  embodied  in  the  following  two  theorems  : 

1.  Multiplying  a  magnitude  by  a  negative  factor, 
multiplies  it  by  the  factor  and  turns  it  in  the  opposite 
direction. 

2.  Multiplying  by  —1  turns  it  in  the  opposite  direc- 
tion without  altering  its  length. 

Note.  When  more  than  two  factors  enter  a  product,  the 
sign  may  be  determined  by  the  theorem,  §  26. 

EXERCISES. 

I.  am  X  ah  x  ac  x  ad.  2.     ax  x  —bx  x  ex  x  dx. 

3.  X  X  —ax  X  —abx  x  —aicx. 

4.  dax  X  —^aW  X  —  ba^mx, 

5.  —lm?y  X  —?tc^y^  x  6ax. 

6.  —2nzn  x  — 57i^x^  x  —7i^yz  —  a^. 

7.  2m  X  n  X  —a  x  —2b. 

8.  —3ax  X  —2km  x  —Hx  x  — 4ibmx» 

9.  —ny  X  gy  X  —2yx  Sbm, 

10.  xy  X  2y^  x  y^x  x  2ayxK 

11.  by'^  X  —^gy  X  —2z^  X  —axh. 

12.  bax  X  anx  x  ^z  x  b'^xy. 

13.  —4J)z  X  —xz  X  —yz  X  agz, 

14.  2c^n  X  2xh  X  —z^  X  —bgz\ 

15.  —e^x  X  3a:  X  eb^  x  ay* 


44  ALQEBRAIG   OPERATIONS, 

i6.  —2e  X  —2^  X  a  X  Ix. 

17.  —  4a.r  X  ^ay  x  —2a?y  x  — i?^^. 

18.  a'^:r  X  —  <^,V^  X  ax^  x  —x^y, 

19.  a:?;2  X  _y2  X  —1  X  ^ax  x  — «V 

20.  wi^^  X  —n^x  X  —mn^  x  w.^  x  —  m^. 

21.  —  a&2;  X  —ay'^  x  ^^i-'  x  a^x^. 

22.  ^.^2  X  ^^^  X  xy  X  —  flfiz:. 

23.  abc  X  — <:?^  x  aa;^  x  —1  X  'dax. 

24.  -ri^:?:  x  ^cx  X  —  T^mx  x  — 4?/2  x  6m. 
4  2  "^ 

2K,     —6mx  X  —2n^x  x  -^ac  x  — -m^. 

DO 

26.  —a  X  ^c  X  — 1  X  T  X  3a2  X  ^xy  x  y. 

27.  —1  X  ax  X  a^x  x  a^x^  x  hx  x  d, 

28.  —an  X  2am^  x  —dmn  x  6n^y  x  —m, 

29.  —mx  X  nx  X  —mn  x  —xy  x  —1. 

30.  —2px  X  —dqx  X  -m^x  x  -^y^  x  — 1. 

bo 

Products  of  Polynomials  by  Monomials. 

73.  The  rule  for  multiplying  a  polynomial  is  given  by  the 
disfcributive  law  (§  65). 

Rule.  Multiply  each  term  of  the  polynomial  hy  the 
monomial,  and  talce  the  algebraic  sum  of  the  products. 

Exercises.    Multiply 

1.  dx^  —  4:xy  —  6y^  by  —  4ax, 

Ans,  —  12ax^  +  IGax^y  +  20axyK 

2.  3.^2  —  xy  -\-  y^  by  3.r. 

3.  x^  -\-  xy  -}-  y^  by  3x,  4.  ax  -{-  by  +  cz  by  axyz, 
5.  dax^—6ay^—7  by  dabx,  6.  4m^  —  6nq  by  —  3mq. 
7.     o^y  —  7«y  —  la'^y  by  Sab, 

74.  The  products  of  aggregates  by  factors  are  formed 
in  the  same  way,  the  parentheses  being  removed,  and 
each  term  of  the  aggregate  multiplied  by  the  factor. 


MULTIPLICATION,  45 

Example.     Clear  the  following  expression  of  parentheses : 

am  {a  —  h  +  c)  —p  \a  —  {Ji  ^  h)  —  m  {a  —  b)]. 
By  the  rule  of  §  73,  the  first  term  will  be  reduced  to 

ah7i  —  amd  +  amc.  (1) 

The  aggregate  of  the  second  term  within  the  large  paren- 
theses will  be 

a  —  h  -\-  k  —  m  {a  —  V) 

=:  a  —  h  -{-  k  —  ma  +  mh^  (2) 

because,  by  the  rule  of  signs  in  multiplication, 

—  m{a—'b)  =  —m  x  a  —  m  x  —h  =  —  ma  +  mb. 

Multiplying  the  sum  (2)  by  —p  and  adding  it  to  (1),  we 
have  for  the  result  required: 

a^m  —  atnb  +  amc  —  pa  -^  pli  —  pk  -\-  pma  —pmb, 

EXERCISES. 

Clear  the  following  expressions  of  parentheses  : 

1.  p{a  -{-  m  —  p)  -{-  q{b  ■—  c)  —-r  {b  +  c). 

2.  {rn  —  an)  x  —  {m  +  an)  y  +  {a7i  —  m)  z, 

3.  a{x  —  y)c  —  b{x  —  y)  d+f{x  +  y)  cd. 

Here  note  tliat  the  coefficient  of  ai  —  y  in  the  first  term  is  ac. 

4.  am  [x  —  a{b  -^  c)]  —  bn  [ax  -}-  b{c  +  d)], 

5.  p  [—  a  {m-\-n)  +  b{m—n)]  —  q[b  {m  —  n)—a{m  +  n)], 

6.  dx  (2q  —  7ic)  +  2y  {6x  —  3c)—z  (2m  +  7n), 

7.  am  [m  {a  —  b)c  —  3h  {2k  —  4:d)  +  4/^]. 

8.  2pq  [da  --^b  —  ec—pq  (2m  —  3^z)]. 

9.  bn  [—'7a  —  n{a-^c)  —  {3  —  a  —  b)], 
10.  p{q  —  r)-{-q{r  —  p)  +  r{p  —  q). 

75.  The  reverse  operation,  of  summing  several  terms  into 
one  or  more  aggregates,  each  multiplied  by  a  factor,  is  of  fre- 
quent application.     Thus,  in  §  65,  having  given 

mp  -{•  mq  -\-  mr, 

we  express  the  sum  in  the  form 

7)2{p  -i-  q  -{-  r). 


46  ALGEBRAIC   OPERATIONS. 

The  rule  for  the  operation  is 

If  the  sum  of  several  terms  having  a  commjon  factor 
is  to  he  formed,  the  coeffieients  of  this  factor  may  he 
added,  and  their  aggregate  he  multiplied  hy  the  factor, 

Note.     This  operation  is,  in  principle,  identical  with  that  of  §  55. 

EXAMPLES. 

ahx  —  lex  —  ady  -\-^dy—  ohx  ■i-4:ady-i-  my — amy  —  dcmx  +  hinx. 

Collecting  the  coefficients  of  x  and  y  as  directed,  we  have 
{ab  —  hc  —  dh  —  dcm  +  bm)  x  -\-  {—ad-\-3d+4rad-}-m—am)  y. 

Applying  the  same  rule  to  the  terms  within  the  parentheses, 

we  find  • 

ab  —  I}c  —  3b  =  h{a  —  c  —  3). 

—  3cm  +  bm  =  m.(b  —  3c). 

—  ad  -\-M  -\-  4:ad  =  Sad  +  3d 

=  {3a  +  3)d 

=  3{a-}-l)d. 

m  —  am  =  m  (1  —  a). 

Substituting  these   expressions,   the    reduced    expression 
becomes 
^h{a  —  c  —  3)+  771  (b  —  3c)'\  X  -^  \3  {a  -{- 1)  d  +  m(l  —  «)]  y. 

The  student  should  now  be  able  to  reverse  the  process,  and 
reduce  this  last  expression  to  its  original  form  by  the  method 

of  §  74. 

EXERCISES. 

In  the  following  exercises,  the  coefficients  of  y,  z,  and 
their  products  are  to  be  aggregated,  so  that  the  results  shall 
be  expressed  as  entire  functions  of  x,  y,  and  z,  as  in  §  55. 

1.  ax  -\-  bx  —  3ax  +  3bx  +  6a:  —  7:r. 

Ans.  (—  '^a  -^  ^b  —  l)x. 

2.  my  -\-  py  —  my  —  2py  — -  3gy. 

3.  mx  —  ny  -^  px  '-'  gy  H-  rx  —  sy. 

Ans,  {7n  -}-  p  -\-  r)  X  --  (n  +  g  +  s)  y, 

4.  3az  —  y  —  2az  -\-  z  —  az  -\-  y. 


MULTIPLICATION.  47 


5- 

abxy  —  Icxy  -f-  J^iz/. 

6. 

d^abxy  —  24rc  —  a^^;  —  7a;jf. 

7. 

^y  —  ^y  —  ^^^^  —  ^^^^  +  ^'^« 

8. 

«m«/  —  Z>/wy  +  a7iy  —  hny. 

9- 

prz  —  "Iqrz  —  ^ppz  +  Sqhz. 

lO. 

cnx  -{-  b7ix  —  amy  —  'Zhny. 

76.  An  entire  function  of  two  quantities  can  be  regarded 
as  an  entire  function  of  either  of  them  (§§  49,  50),  and  when 
expressed  as  a  function  of  one  may  be  transformed  into  a  func- 
tion of  the  other. 

Example.     The  expression 

{U  +  3)  a;3  _  (4^2  _  ^a)  x^  +  {d^  -^2a  +  l)x  —  a^ 

has  the  form  of  an  entire  function  of  x.    It  is  required  to 
express  it  as  an  entire  function  of  a. 
Clearing  of  parentheses,  it  becomes 

2ax!^  -\-  dx^  —  4oa^x^  +  2ax^  +  a^x  —  2ax  -\-  x  —  a^. 

Now,  collecting  the  coefficients  of  a^,  a^,  etc.,  separately,  it 
becomes 

(_  4:x^  ^x  —  1)  a^  +  {2x^  +  2x^  —  2x)  a  +  dx^  -}-  x, 
which  is  the  required  form. 

EXERCISES. 

Express  the  following  as  entire  functions  of  y : 

1.  {dy^—4.y)x^  +  {y^—2y^-}-l)x^-{-{2y^-^6y^—7)x—y^—G. 

2.  {y^  —  y^)  x^  +  {y^  —  y)^  i-  y^  —  1- 

3.  (\/5  —  2y^)  x^  +  (y4  _  2y^)  x^  +  {y^  —  2y)  x  •{- y'^  —  2, 

4.  {y'  +  3^/2)  ^  +  (2/4  +  3f/3)  x^  +  (^3  +  32/)  x^  +  (^/2  +  3)  x. 

Multiplication  of  Polynomials  by  Polynomials. 

77.  Let  us  consider  the  product 

{a-{-l)){p  +  q^  r). 
This  is  of  the  same  form  as  equation  (1)  of  §  65,  {a  -{-  h) 
taking  the  place  of  m.     Therefore  the  product  just  written  is 
equal  to 

(«  +  ^)  ^  +  («  +  ^)  ^  +  {ct  +  h)  r. 


48  ALGEBRAIC   OPERATIONS, 

But  {a  +  l))p  :=z  np  -{-  bp, 

(a  -^  b)  q  =  aq  -{-  bq. 
(a  +  b)  r  =^  a?'  +  br. 
Therefore  the  product  is 

ap  -\-  bp  -\-  aq  -\-  bq  -{-  ar  +  br. 

It  would  have  been  still  shorter  to  first  clear  the  paren- 
theses from  {a  +  b),  putting  the  product  into  the  form 

(^{P^  q  -\-r)  -\-b{p  -\-q  +  r). 

Clearing  the  parentheses  again,  we  should  get  the  same 
result  as  before. 

We  have  therefore  the  following  rule  for  multiplying  aggre- 
gates : 

78.  Rule.  Multiply  each  term  of  the  inidtiplicancl 
hy  each  term  of  the  multiplier,  and  add  the  products 
ivith  their  proper  algebraic  signs, 

EXERCISES. 

1.  (a  +  b)  {2a  —  bn^  —  2bn^), 

2.  [a  —  b)  {^m  -f  271  —  bahnn), 
3-     (y)i^  —  ^2)  {2mn  +  ^^m  +  qn), 

4.  {p'^  +  q^  -\-  r^)  {pq  +  qr  -f-  rp). 

5.  (2a  —  db)  {2a  f  2b). 

6.  {)72x  —  ny)  {mx  +  ny), 

79.  It  is  frequently  necessary  to  multiply  polynomials 
containing  powers  of  the  same  letter.  In  this  case  the  begin- 
ner may  find  it  easier  to  arrange  multiplicand,  multiplier,  and 
product  under  each  other,  as  in  arithmetical  multiplication. 

Ex.  I.     Multiply  Ix^  —62:2  +  5^  —  4  by  ^x^  —  Ax  —  5. 

The  first  line   under  work. 

tlie    multiplier     contains  7.^'^ — 6x^-j-6x — 4 

the  products  of  the  sev-  0^2 A^^  5 

eral  terms  of  the  multipli- 

candby3aj2      The  second  21:^5  — 18.^4  _|_  15.^3 _i2:z;2 

contains  the  products  by  — 28^;^  + 24^;^ — 20rt^2_j_i(5;;p 

-4x,  and  the  third  by  -5.  _  35^3  _^  30^2_  25^7;  +  20 

Like    terms    are     placed — 

under  each  other  to  facil-  21x^—4:6x^+    4:X^—    2x^-^    92:  +  20 
itate  the  addition. 


MULTIPLICATION.  49 


Ex.  2.     Multiply  m  -\-  nx  -\-  px^  hj  a  —  hx, 
a  —  hx 


am  +  anx  +  a'px^ 

—  hmx  —  hnx^  —  hpoi^ 


am  +  {an  —  bm)  x  -f-  {ap  ~  hn)  ^  —  hpo? 

In  the  following  exercises  arrange  the  terms  according  to 
^.ne  powers  and  products  of  the  leading  letters,  a^  h,  x,  y,  or  z 

Multiply 

1.  3^2  +  5«  +  7  by  2^2  _  3a  +  4. 

2.  a^  -\-  ah  +  y^  hy  a  —  K 

3.  a^  -\-  a^  -\-  ax^  +  ^^  by  ^  —  x. 

4.  a^  —  a^  '\-  a  —  1  by  ^^  —  a  -\-  1, 

5.  x^  -\-  ao^  -\-  a^x^  +  a^x  ■\-  a^  ]dj  x  —  a. 

6.  a  -\-  hz  -\-  cz^  +  f/2;^  by  m  —  nz  -\-  pz\ 

7.  3«2  f  5«  +  7  by  2«2  +  3^  —  4. 

8.  a2  _  ^^  ^  ^2  by  ^  ^  J. 

9;  a^  4-  a^o;  +  ax^  -\-  a^  hj  a  —  x, 

10.  a^  —  a^  +  a  —  1  hj  a^  -^  a  —  1. 

11.  x/^  -{-  ao(^  -^  a^x^  +  aH  -^  a^  hj  x  +  a. 

12.  a  -\- Iz  -{-  cz^  -{-  dz^  by  m  +  ^^2;  —pz\ 

13-  (^  +  ^''^)  (^  +  ^^)- 

14.  (a  +  Z^:c  +  c:?:2)  i^'jfyi  j^  fix  ^  px^), 

15-  (y«  -  3^  +  2)  (*/2  -  2). 

16.  (2/8  +  2/3  +  y  +  1)  (2/3  +  y  +  1) 

17-  («/'  -  2«/3  +  3?/  -  4)  (^3  +  22/2  +  3?/  +  4). 

18.  Sft^^a;  —  3a2y  +  2a2»  by  a^  —  a" 

19.  a?  +  6«5  +  -5  by  «  —  ~i. 

20.  («  4-  ^)  4-  (a  —  i)  by  («  4-  ^)  —  (a  —  ^). 

21.  a2  —  ^2  _j.  (^  „  ^)  by  «2  ^  Z>2  ^  (a  4-  Z^). 

22.  «  4-  ^  -f-  ^  by  ^  —  ^  4-  c. 

23.  ^2  +  ^2  _  (3^2  _^  j2)  by  2a  +  2^  —  2  (aj  —  ^). 

24.  2{a  ^h)  -\-  X  ~  y  hy  a  '\-  h  —  {x  -{-  y), 

25.  «:^"*  4-  hx"*  —  «^^:c  by  aa;2  4-  ba^, 

26.  «'"  —  b''  by  a"*  4-  ^". 

4 


50  ALGEBRAIC    OPERATIONS. 

27.  —  Ibx^y  +  ^xy'^  —  12^^  by  —  hxy. 

2  7  1 

28.  -x^  +  3«.?;  —  ~a^  by  20;^  —  aj:?;  —  -al 

3  D         ^  4 

ISToTE.  Aggregates  entering  into  either  factor  should  be 
simplified  before  multiplying. 

Special  Fornis  of  Multiplication. 

80.  1.  To  find  the  square  of  a  binomial,  as  a  -{-l.  We 
multiply  «  +  ^  by  a  +  Z>. 

a  {a  -\-  d)  =  a^  -\-  ab 
b{a  +  b)  =  ah  +  l^ 

{a  +  h){a^l))  =  o?  +  "^ah  +  W' 
Hence,  {a  +  Vf  =  «2  +  %ab  +  ^2  (1) 

2.  We  find,  in  the  same  way, 

{a  —  ly  =  a^  —  2ab  +  P,  (2) 

These  forms  may  be  expressed  in  words  thus: 

Theorem.  The  square  of  a  binomial  is  equal  to  the 
sum  of  the  squares  of  its  two  terms,  plus  or  minus  twice 
their  product. 

3.  To  find  the  product  of  a  -f-  ^  by  a  —  b, 

a{a  -{-  b)  =  a^  -{-  ab 
-^b(a  +  b)  =      -ab-W 
Adding,    {a  +  b)  {a  —  b)  =  a^  —  b\  (3) 

That  is . 

Theorem.  The  product  of  the  sum  and  difference  of 
two  numbers  is  equal  to  the  diflference  of  their  squares. 

The  forms  (1),  (2),  and  (3)  should  be  memorized  by  the  student,  owing  j 
to  their  constant  occurrence.  ^ 

When  Z>  =  1,  the  form  (3)  becomes 

(a  +  1)  (a  -  1)  =  6^2  —  1. 

The  student  should  test  these  formulae  by  examples  like 
the  following : 

(9  +  ^f  =  92  +  2.9.4  H-  42  =  81  +  72  -f  16  rrr  169. 
(9  _  4)2  =rr  92  —  2.9.4  -I-  42  =:  81  -  72  -f  16  =  25. 


2. 

(m  — 

-  2w)2. 

4- 

(ix- 

-5yy. 

6. 

(3x  +  1)  (3x  - 

■!)• 

8. 

(5a^5 

-  3)  (5a;3 

+  3) 

MULTIPLICATION.  51 

(9  +  4)  (9  —  4)  r=  92  —  4-  =  65. 

Prove  these  three  equations  by  computing  the  left-hand 
member  directly. 

EXERCISES. 

Write  on  sight  the  values  of 

I.  {m  -\-  2^)2. 

3.  {da  -  2b)\ 

5.  {%x  +  y)  {%x  -  y). 
7.  {4ic2  +  1)  {4rX'^  —  1). 

81.  Because  the  product  of  two  negative  factors  is  positive, 
it  follows  that  the  square  of  a  negative  quantity  is  positive. 

Examples.      {— of  —  a^  =.  {+  a)\ 

{b  -  ay  =  a^  —  2ab  +  b^  =  (a  -  by. 

Hence, 

27ie   ejopression  a^  —  2ab  +  b^  is   the  square  hoth  of 
a  —  b  and  of  b  —  a, 

83.  We  have  ^  a  x  a  :=  —  a^. 

Hence, 

The  pj^oduct  of  equal  factors  with  opposite  signs  is  a 
negative  square. 

Example,       —  {a  —  b){a  —  b)  =  —  a^  +  2ab  —  b% 

which  is  the  negative  of  (2).     Because  —  {a  —  b)  =  b  —  a, 
this  equation  may  be  written  in  the  form, 

{b  —  a){a  —  b)  =  —  a^  +  2ab  —  i^, 

which  is  readily  obtained  by  direct  multiplication. 

EXERCISES. 

Write  on  sight  the  values  of 
I.     —  {a  -i-  b)  X  —  {a  +  b). 

2.    (^  —  y){y  —  ^)-  3-    {^  -\-y){-^  —  y)- 

4.  {2a  —  3b)  (3b  —  2a).  5.     (3^  —  2a)  {—  3b  +  2a). 

6.  (am  —  bn)  {bn  —  am).         7.     {xy  —  2)  (2  —  xy). 


52  ALGEBRAIC   OPERATIONS, 

CHAPTER    III. 

DIVISION. 

83.  The  problem  of  algebraic  diyision  is  to  find  such  an 
axpression  that,  when  multiplied  by  the  divisor,  the  product 
shall  be  the  dividend. 

This  expression  is  called  the  quotient. 

In  Algebra,  the  quotient  of  two  quantities  may  always  be 
indicated  by  a  fraction,  of  which  the  numerator  is  the  divi- 
dend and  the  denominator  the  divisor. 

Sometimes  the  numerator  cannot  be  exactly  divided  by  the 
denominator.  The  expression  must  then  be  treated  as  a  frac- 
tion, by  methods  to  be  explained  in  the  next  chapter. 

Sometimes  the  divisor  will  exactly  divide  the  dividend. 
Such  cases  form  the  subject  of  the  present  chapter. 

Division  of  Monomials  by  Monomials. 

84.  In  order  that  a  dividend  may  be  exactly  divisi- 
ble by  a  divisor,  it  is  necessary  that  it  shall  contain  the 
divisor  as  a  factor. 

Ex.  I.     15  is  exactly  divisible  by  3,  because  3-5  =r  15. 
2.  The  product  a¥G  is  exactly  divisible  by  ac,  because  ac  is 
a  factor  of  it. 

To  divide  one  expression  by  another  which  is  an  exact 
divisor  of  it: 

EuLE.  Remove  from  the  dividend  those  factors  the 
product  of  ivhich  is  equal  to  the  divisor.  The  remain- 
ing  factors  luill  be  the  quotient, 

85.  Rule  of  Exponents.  If  both  dividend  and  divisor 
contain  the  same  symbol,  with  different  exponents,  say  m  and 
fly  then,  because  the  dividend  contains  this  symbol  m  times  as 
a  factor,  and  the  divisor  yi  times,  the  quotient  will  contain  it 
m  —  n  times.     Hence, 


DIVISION.  63 

In  dividing,  exponents  of  like  symbols  are  to  he  sub^ 
traeted. 

EXERCISES. 

1.  Divide  26.t«/  by  2^.  Ans,  l^x. 

2.  Divide  21^2^,^  by  Ihc. 

3.  Divide  o(^  by  x^,  Ans.  x, 

4.  Divide  ISa^  by  6«.  Ans,  3a. 

5.  Divide  15a^m  by  3^^.  Afis.  bam. 

6.  Divide  14«3m2  by  7fi^m. 

7.  Divide'  16^^^^  by  Sahn^.  8.  Divide  36a:i/22;3  by  Qxyz. 
9.  Divide  40a2a;8^5  by  lOa^:^^^.  10.  Divide  3ba}p  by  Ta^'. 

Rule  of  Signs  in  Division. 

86.  The  rule  of  signs  in  division  corresponds  to  that  in 
multiplication,  namely : 

//  dividend  and  divisor  have  the  same  sign,  the  quo- 
tient  is  positive. 

If  they  have  opposite  signs,  the  quotient  is  negative. 

Proof. 

■\-mx  -V-  (  +  m)  =  +x,  because  +x  x  {-\-m)  =  -\rmx.  . 

-\-mx  -~  (— '/^O  =  —^9  "  —^  X  (—^^0  =  +^?^^- 
^mx  -=-  (  +  m)  =  —X,  "  —X  X  (  +  W)  =  —mx. 
—mx  -^  {—m)  =  -\-x,        "         -\-x  X  (— m)  =  —7nx. 

The  condition  to  be  fulfilled  in  all  four  of  these  cases  is 
that  the  product,  quotient  x  divisor,  shall  have  the  same  alge- 
braic sign  as  the  dividend. 


EXERCISES. 

Divide 

I..    -\-a  by  -f-  a. 

Ans.  +  1. 

2.      +  «5  by  —  a. 

Ans.  —  1. 

3.     —  a  by   +  a. 

Ans.  —  1. 

4.     —ah\  —a. 

Ans.   +  1. 

5.     _  33ahnx  by 

llax. 

Ans.  —  dam. 

6.     —  24:X^yz  by  12xyz. 

Ans.  —  2x. 

7.     21ani^x'"  by  - 

-  7amx"', 

A 

ns.  —  3m.T"'~". 

54 


ALGEBRAIC   OPERATIONS. 


8. 
9 

lO 

II 

12 

13 
14, 

16 
17 


—  ISa^p^  by  —  6^^^.  ^/is.  3a'^~'*jy'^-i. 

—  16a^xiny^  by  4aa;2?/'*. 
14^^^  by  —  7by^, 

—  lUH^'lC^  by  —  ^lH^h\ 
V^Ka  —  Vfd'  by  3(^-Z>)2c. 
42  (2;  —  2/)^  by  —  7  (x  —  yy. 

—  44a*  {x  —  yY  by  lla^  {x  —  yy. 

—  45^^  {a  —  Z>)^  by  9Z/^  {a  —  Z>)^. 

—  48  {m  +  7z)^  by  —  8  (^7?  +  oi)^, 
64  (a  +  ly  {x  —  ij)^  by  4  (a  +  I)  {x  —  y). 


Ans.  4z{a  —  i)  c^. 


Division  of  Polynomials  by  Monomials. 

8*7.  By  the  distributive  law  in  multiplication^,  whatever 
quantities  the  symbols  m,  a,  b,  c,  etc.,  may  represent,  we  have : 

{a  +  b  -^  c  -{•  etc.)  x  m  =  7na  -\-  mb  -\-  mc  +  etc. 
Therefore,  by  the  condition  of  division, 

{ma  +  7nb  +  mc  +  etc.)  -^m=za-\-b-\-c-\-  etc. 
We  therefore  conclude, 

1.  In  order  that  a  polynomial  may  be  exactly  divisi- 
ble by  a  monomial,  each  of  its  terms  mnst  be  so 
divisible. 

2.  The  quotient  will  be  the  algebraic  sum  of  the 
separate  quotients  found  by  dividing  the  different  terms 
of  the  polynomial. 

EXERCISES. 

Divide 

1.  2fl^2  _|.  6^3^  _  8^5^2  }3y  2a\     Ans.  1  4-  ^ax  —  4a% 

2.  Qm^n  —  12m^/^2  —  ISmn^  by  Qmn, 

3.  ^a%^  —  16^4^,4  +  8a5J3  by  4.aW. 

4.  ^xy^  —  Sa^y^  +  4x^y  by  -—  4:xy. 

5.  12abx  —  24:abx^  by  —  Uabx, 

6.  21am^x:^  —  Ua^m^x^  +  2Sa^7n^x^  hy  —  7amx!^, 

7.  '72a^x  +  24^:^  +  A8ax^  by  24^:?;. 

8.  a{b  —  c)  -}-  b  (c  —  a)  -{-  c  {a  —  b)  ■\-  abc  by  a^^c. 

9.  27  («  -  Z')5  _  18  («  -  by  +  9{a~  by  by  9  (a  -  b). 
10.  a^  {a  —  by  —  ci^  («  —  by'  by  a''  {a  —  Z>)'^. 


Divisioif,  55 

11.  {a  +  by  {a  -  hf  +  {a-^-hf  {a-hy  by  {a  +  l)  {a-h). 

12.  10  (x  +  y)^{x  -.yY-.^[x  ^  yy  (x  —  yf 

by  6{x  -\-  y)  (x  —  y). 

13.  («  +  Z^)  (a  -  Z^)  by  a^  -  ^l 

Factors  and  Multiples. 

88.  As  in  Arithmetic  some  numbers  are  composite  and 
others  prime,  so  in  Algebra  some  expressions  admit  of  being 
divided  into  algebraic  factors,  while  others  do  not.  The  latter 
are  by  analogy  called  Prime  and  the  former  Composite. 

A  single  symbol,  as  a  or  x,  is  necessarily  prime. 

A  product  of  several  symbols  is  of  course  composite,  and 
can  be  divided  into  factors  at  sight. 

A  binomial  or  polynomial  is  sometimes  prime  and  some- 
times composite,  but  no  universal  rule  can  be  given  for  dis- 
tinguishing the  two  cases. 

89.  When  the  same  symbol  01  expression  is  a  factor  of  all 
the  terms  of  a  polynomial,  the  latter  is  divisible  by  it. 

EXAMPLES. 

I.     ax  -{-  dbx^  +  c^GX^  ^=1  a  (x  -t-  Ix^  -\-  «c.t^). 
c^h^x  +  c^Wx^  =r  aWx  {h  +  ax^. 


2 


EXERCISES. 

Factor 

I.     ax^  -f  c^x,  2.     a^'^cy  +  c^hc^y  +  ahh^y. 

3.     a^^  h^  +  a'^  h'^'^.  4.     a^^  x^  —  a^^  x^^  +  a^  x^^. 

5.     a^i^^c^^  -(-  a^^b^^c^  +  a^^b^c^^, 

90.  There  are  certain  forms  of  composite  expressions 
which  should  be  memorized,  so  as  to  be  easily  recognized. 
The  following  are  the  inverse  of  those  derived  in  §  80. 

1.  a^  +  2ab  -^  b^  =  {a -\-  b)\ 

2.  a}  —  2ab  -{-  b^  =:  (a  —  b)\ ' 

3.  a'-b'^  =  {a  +  b)  {a  -  b). 

The  form  (3)  can  be  applied  to  any  difference  of  even 
powers ;  thus, 


56  ALQEBBAIG   OPERATIONS, 

a^-b^=.  {a^  +  h^)  (^2  _  ^2)  . 

and,  in  general,    a^'^  —  W^  —  {a''  +  l'^)  (a''  —  b""). 

If  the  exponent  is  a  multiple  of  4,  the  second  factor  can  be 
again  divided. 

EXAM  PLES. 

a^-h^  ^  (a?  +  h^)  («2  -  ^2)  ^  {d^j^W)  (a-{-h)  (a-b). 

a?^¥  =  {a'  +  ¥)  {a^  -  ¥)  =  {a'  +  ¥)  (a^  +  b^)  (a-^b)  (a^b). 

When  b  is  equal  to  1  or  2,  the  forms  become 

a^  —  l  =z  {a  +  l)(a  —  1). 
a^-.4c  =  {a  +  2)  {a  —  2). 
a^-^2a  +  l  =  {a  +  1)K 

a^-2a  +  l  =  {a-lY  =  {1-  a^. 
a^  —  4^.  +  4  =  {a  —  2)2  =  (2  —  ay. 

By  putting  2b  for  b,  they  give 

a2  _  4^2  =  (a  +  2b)  (a  —  2Z>). 
^2  +  4a^  +  4Z>2  =  (^  ^  2b)\ 

EXERCISES. 

Divide  the  following  expressions  into  as  many  factors  as 
possible  : 

1.  x^  —  16.  Ans.  (^2  4-  4)  (a:  +  2)  (x  —  2). 

2.  y^  —  16a;4. 

3.  a;2  +  6:r  +  9.  Ans.  (:r  +  3)2. 

4.  a;2  _  6:r  -f  9. 

5.  4.a^x^  —  Wy\  6.     lea^a;^  __  1, 

7.     9:^2  __  i2:r^  -f-  4^2.  8.     ^2:^2  +  2«a;y  +  y\ 

9.       4fl^2^2  ^  ^^^bxy  +  62i^^l  10.       «4  +  4«2^  ^  4J2. 

11.  x!^  —  2.Ty  _|.  ^4^  1 2.     a;4  _  4^2^2  ^  4^4^ 

13.  ^4  _  4^2^2  ^  4^4;  14.     a^--a%\ 

15.  ^2/1  _  2^w  4.  1.  16.     x^  —  4aa;^  +  4:a\ 

17.  1  —  2/*- 

18,  2:6^  +  2x^yh  +  t/^js;.  • 

^725.    Z{X^  +  2xhf  +  y^)    =  ^  (aj3  4.  ^8)2. 


DIVISION, 

19. 

21. 

23- 
25- 

20. 
22. 
24. 
26, 

^2m  _  ^» 
4^^  —  9itY. 

0^  _  i^;2^6^ 

2:4m  „  2:6^  ^  1. 

27. 

a;2  +  a;  +  -• 

28. 

^2m  ^  ^  ^ 

4 

57 


91.  By  combining  the  preceding  forms,  yet  other  forms 
may  be  found. 

For  example,  the  factors 

(«2  +  ah-\-  b^)  (a2  -  aJ  +  b^),  (1) 

.  are  respectively  the  sum  and  difference  of  the  quantities 
Or  +  W-    and    ab. 

Hence  the  product  (1)  is  equal  to  the  difference  of  the 
squares  of  these  quantities,  or  to 

(«2  ^  52)2  _  aW  =  a^  +  aW  +  ¥, 
Hence  the  latter  quantity  can  be  factored  as  follows : 
a*  +  aW  ■\-¥  —  {a^  +  ab  +  b^)  {a^  -^  ab  ■\-  b'^). 

EXERCISES. 

Factor 

I.     x^  +  xY  -\-  y^'  2.     a^  +  8«2^2  ^  iej4, 

3.     «4  +  9a2a;2  +  Sla;^.  4.     a^""  ^  d^^  b^^ -\- ¥^, 

8.  m^  —  a^  _|_  2a^  —  ¥,     Ans,  {m  —  a  -\-  b)  {m  -\-  a  —  b)* 
Here  the  last  tliree  terms  are  a  negative  square.     Compare  §  82. 

9.  a^  —  4^2  _|_  4^^  —  ^^         10^     ^8  —  4^j2  _|_  4^j^  —  fjf.2^ 

93.  The  following  expression  occurs  in  investigating  the 
area  of  a  triangle  of  which  the  sides  are  given : 

(a  ^  b  -\-  c)  {a  ^  b  —  c)  {a  —  b  -\-  c)  {(I  —  b  —  c),        (1) 

By  §  80,  3,  the  product  of  the  first  pair  of  factors  is 

(a  +  Z>)2  —  c2  =r  a2  4-  2«&  +  Z>2  _  c2  ; 

and  that  of  the  second  pair, 


68  ALOEBRAIG   OPERATIONS. 

By  the  same  principle,  the  product  of  these  products  is 

(^2  4-  2>2  _  ^2)2  _  ^aW, 

which  we  readily  find  to  be 

^4  _|_  ^4  _!_  ^4  _  2^2^  _  2Z»%2  _  2c^a\  (2) 

Hence  this  expression  (2)  can  be  divided  into  the  four 
factors  (1). 

Factors  of  Binomials. 

93.  Let  us  multiply 
x""-^  +  ax""-^  4-  a^x^~^  + +  a'^'^^x  +  a^-i  by  x  —  a. 

OPERATION. 

a;  —  a 

it'«  _j_  ax^-^  +  «2^w-2  +  a%^-3  4_ 4_  ^/^-i  ^ 

—  ax'^-'^  —  a^a;'^-^  —  ahf'-^  —  ....  —  fl^^^-i^.  —  a^ 
Prod.,  ii;^        0  0  0  0     —  a^ 

The  intermediate  terms  all  cancel  each  other  in  the  product, 
leaving  only  the  two  extreme  terms. 

The  product  of  the  multiplicand  by  .^;  —  a  is  therefore 
x^  _  a'^.  Hence,  if  we  divide  x'^  —  a'^  by  x  —  a,  the  quotient 
will  be  the  above  expression.  Hence  the  binomial  x'^  —  a^ 
may  be  factored  as  follows : 

x'>^  -^aP'  =  {x  —  a)  {x'^-^  +  ax'^-^-i-a^x^-^+ -\-a'^~^x-\-a^-^). 

Therefore  we  have, 

Theorem.  The  difference  of  any  power  of  two  num- 
Tbers  is  divisible  by  the  difference  of  the  numbers 
themselves. 

Illustration.  The  difference  between  any  power  of  7 
and  the   same  power  of  2   is  divisible  by  7  —  2  =  5.    For 

instance, 

72  _  22  ^      45  ^  5.9. 

73  —  23  =    335  =  5.67. 
74 -2^  =  2385  =  5.477. 

etc.  etc.         etc 


DIVISION.  59 

94.  Let  us  multiply 

^71-1  _  axr^-"-  +  a^x^-^  — +  (—  «)'^-2^  +  (—  a)^-i 

bj    a;  +  a  =  i^;  —  (—  ^). 

Eem.  This  expression  is  exactly  like  the  preceding,  except 
that  —  a  is  substituted  for  a.  It  will  be  noticed  that  the 
coefficients  of  the  powers  of  x  in  the  multiplicand  are  the 
powers  of  —  a,  because 

{—ay  =  —a, 

(_a)3  ^  ^a% 

{-ay  =:  ^aS 

etc.  etc. 

The  sign  of  the  last  term  will  be  positive  or  negative, 
according  as  n  —  1  is  an  even  or  odd  number. 

OPERATION. 

xn-^—ax'^'^  +  a^x'^-^—aH^'^-^...,-\-{-'aY-^X'\-{—ay-^ 
X  -{-  a  :=  X  —  {—  a) 

xn  —  ax""-^  -h  o?x^-^  —  aH'^-^  . . .  +  {—ay-^ 

+  ax'^-^  —  a^x^-^  +  a^x^-^.,..  —  {—a)^-^x  —  (— r/)^ 

Prod.,  :r^  0  0  0  0  —  {—aY' 

Ik 

f^'  The  multiplier  x  -[-  a  is  the  same   as   x  —  {—  a)    (§  59). 

In  multiplying  the  first  terms,  we  use   +  a,  and  in  the  last 

ones  —  (—  a),  because  the  latter  shows  the  form  better. 

Hence,   reasoning  as  in  (I),  the  expression  x^  —  (—  a)^ 

admits  of  being  factored  thus  : 

a^^  —  (—  a)""  =  {x  +  a)  [x"^'^  —  ax"^'^  +  a^x^'^  — 

....  +  (—  a)^-^x  4-  (~  a)^-i]. 

If  n  is  an  even  number,  then  (—  a)^  =  a%  and 

xn  _  (—  a)^  —  x^  —  a^. 

If  n  is  an'  odd  number,  then  {—  a^)  =z  —  a%  and 

xn  —  (_  aY  =  x^  +  a^ 
Therefore, 

Theorem  1.  When  n  is  odd,  the  binomial  af'-^-a^  is 
divisible  by  riJ+ a. 


60  ALGEBRAIC   OPERATIONS. 

Theorem  2.  When  n  is  even,  tlie  binomial  x^—a"^  is 
divisible  'by  x-\-a. 

Note.  These  theorems  could  have  been  deduced  imme- 
diately from  that  of  §  93,  by  changing  a  infco  —  a,  because 
X  —  a  would  then  have  been  changed  to  x  -{-  a,  and  x'^  —  a^ 
to  x^  -f-  a'^  or  x'^  —  a^,  according  as  7i  was  odd  or  even. 

The  forms  of  the  factors  in  the  two  cases  are  : 

When  n  is  odd,   • 
x'^  -\-  a""  =  {x  +  a)  {x""-^  —  ax'^-'^  +  a^x^~^  — . . . .  -f  a^-i). 

When  n  is  even, 
r^n  __an  —  (x  -{-  a)  {x^-^  —  ax^-^  +  a^x^'^  —  . . . .  -a^-^),      (a) 

In  the  latter  case^  the  last  factor  can  still  be  divided,  be- 
cause x^  —  a^  is  divisible  hjx  —  «  as  well  as  by  x  +  a.     We 
find,  by  multiplication, 
{x  —  a)  {x""-^  +  a^x^-^  -f  «4^-6  +  ....  +  «7i-2) 

—  x''-^  —  ax^-^  +  a^x""-^  —  aH""-^  +  ....  +  oP'-'^x  —  oP'-K 

Therefore,  from  the  last  equation  {a)  we  have : 
When  n  is  even, 
x^  —  oP  z=L  (x -\-a)  (x—a)  {x"^'^ -f  a^x^-^-{-  d^x^-^  —  ....  +  op-'^). 

EXERCISES. 

Factor  the  following  expressions,  and  when  they  are  purely 
numerical,  prove  the  results. 

Ans,  (5  +  2)  (5  —  2). 
52  —  22  =r  25  —  4  =  21 ; 

(5  4_2)(5-2)   =  7.3  rr:  21.] 


I. 

5^  —  Jd^. 

VP 

roof. 

(5 

2. 

53  -  23. 

4. 

55  —  25. 

6. 

73  +  23. 

8. 

7*  —  2*. 

10. 

o^-aK 

12. 

x^  -—aK 

14. 

x^  +  ct\ 

16 

%a^  __  2753. 

18. 

^  +  %y\ 

20. 

Sa^  +  27¥. 

3. 

5^  —  2*. 

5- 

56-26. 

7. 

73  —  23. 

9- 

x^  —  a\ 

II. 

x^  -  a\ 

IS- 

x^ -\-a^ 

IS. 

a^  -  8^3. 

17. 

16^4  —  ¥. 

19. 

xi  _  162/4. 

21. 

x^  —  G4a6„ 

DIVISION.  61 

Least  Common  Multiple. 

95.  Def.  A  Common  Multiple  of  several  quanti- 
ties is  any  expression  of  which  all  the  quantities  are 
factors. 

Example.  The  expression  am/^n^  is  a  common  multiple  of 
the  quantities  a,  m,  n^  arriy  amn,  am^,  m^n^,  etc.,  and  finally  of 
the  expression  itself,  mn^nK  But  it  is  not  a  multiple  of  a^,  nor 
of  X,  nor  of  any  other  symbol  which  does  not  enter  into  it  as  a 
factor. 

Def,  The  Least  Common  Multiple  of  several 
quantities  is  the  common  multiple  which  is  of  lowest 
degree.     It  is  written  for  shortness  L.  C.  M. 

KuLE  FOR  eii^di:n^g  the  L.  C.  M.  Factor  the  several 
quantities  as  far  as  possible. 

If  the  quantities  have  no  common  factor,  the  least 
common  multiple  is  their  product. 

If  several  of  the  quantities  have  a  common  factor, 
the  multiple  required  is  the  product  of  all  the  factors, 
each  of  them  heing  raised  to  the  highest  power  which  it 
has  in  any  of  the  given  quantities, 

Ex.  I.    Let  the  given  quantities  be 

2ab,        U\,         6ac. 

The  factors  are  2,  3,  a,  h,  and  c.  The  highest  power  of  h  is 
l^,  while  a  and  c  only  enter  to  the  first  power.     Hence, 

L.  C.  M.  =:  6ab^c. 

Ex.  2.    a^  —  b%  a^  +  2ab  +  b^  a^  —  2ab  +  IP,  a^  —  ¥. 

Factoring,  we  find  the  expressions  to  be, 

(^  +  b){a-^  b\     {a  +  bf,     {a  -  bf,    {a^  +  b^'  (a  +  b){a^  h). 

By  the  rule,  the  L.  C.  M.  required  is 

(a  +  byia-^-bfia^  +  V'). 


62  ALGEBRAIC    OPERATIONS, 

EXERCISES. 

Find  the  L.  CM.  of 

I.  xy,  xz,  yz,  2.     d^i,  ¥c,  cH,  d^CL 

3.  a,  ah,  abc,  abed.  4.     a^,  a¥,  Id^. 

5.  x^  —  y\  X  -^y,  X  —  y. 

6.  ^2  _  4^  ^  _  4^  _l_  4^  x^  +  4rX  -{-  4. 

7.  16a^x^  —  4:771%  2ax  +  m,  2ax  —  m, 

8.  ^2  _  1^  ^2  ^  1^  ics  —  2a;  +  1,  x'^  -^-^x  -\-  1. 

9.  4a  {h  +  c),  h(a  —  c),  2aK 

10.  2(^  —  ^)2,  2(^  +  ^)2,  2  (a  — Z>)(«  + Z>). 

11.  3(:r  +  ^),  ^{x-y),  d{x^  +  y% 

12.  a  —  ^>,    ^2  __  ^2^    ^3  _  ^3^    ^4  _  ^4. 

13-  X  -{-y,  x  —  y,  a  -^1),  a  —  K 

14.  o:^  —  a^,  ic^  +  a^,  rr^  —  c^^  X  -\-  a, 

15.  2^  —  64^6,  x^  —  16aS  x^  —  4^2. 

16.  a-  +  b,  a^  +  2«^  +  b%  a^  —  ¥, 

Division  of  one  Polynomial  by  another. 

If  the  dividend  and  divisor  are  both  polynomials,  and  entire 
functions  of  the  same  symbol,  and  if  the  degree  of  the  numer- 
ator is  not  less  than  that  of  the  denominator,  a  division  may 
be  performed  and  a  remainder  obtained.  The  method  of 
dividing  is  similar  to  long  division  in  Arithmetic. 

96,  Case  I.  When  there  is  only  one  algebraic  sym^ 
hoi  in  the  dividend  and  divisor. 

Let  us  perform  the  division, 

3a;4  _  4^3  +  2ir2  4-  3a:  —  1  -f-  x^  —  x-\-  1. 

We  first  find  the  quotient  of  the  highest  term  of  the  divi- 
sor x^,  into  the  highest  term  of  the  dividend  ^x^,  multiply  the 
whole  divisor  by  the  quotient  dx^,  and  subtract  the  product 
from  the  dividend.  We  repeat  the  process  on  the  remainder, 
and  continue  doing  so  until  the  remainder  has  no  power  of  x 
so  high  as  the  highest  term  of  the  divisor.  The  work  is  most 
conveniently  arranged  as  follows : 


DIVISION. 

Dividend. 

Divisor. 

3^^  _  /^x^  _f  2.^'2  4-  3a;  —  1 

\x^  —  x-i-l 

3a;«  X  Divisor,         '^X^  —  Zx^  +  dx^ 

3a;2  _  ^  _  2 

First  Remainder,            —     CC^  —     X^  -{-  3x  —  1 

—X  X  Divisor,                 —     iK^  +     i^:^  —     X 

Second  Remainder,                       —  2^;^  +  4:X  —  1 

-2  X  Divisor,                                 —2x^-{-2x  —  2> 

Third  and  last  Remainder,                                2^  +  1 

63 


Quotient. 


The  division  can  be  carried  no  farther  without  fractions, 
because  x^  will  not  go  into  x.  We  now  apply  the  same  rule  as 
in  Arithmetic,  by  adding  to  the  quotient  a  fraction  of  which 
the  numerator  is  the  remainder  and  the  denominator  the 
divisor.     The  result  is, 

x^  —  x-{-l  x^—x-\-l      ^  ' 

This  result  may  now  be  proved  by  multiplying  the  quotient 
by  the  divisor  and  adding  the  remainder. 

There  is  an  analogy  between  the  result  (a)  and  the  cor- 
responding one  of  Arithmetic.  An  algebraic  fraction  like  (a), 
in  which  the  degree  of  the  numerator  is  greater  than  that  of 
the  denominator  may  be  called  an  improijer  fraction.  As  in 
Arithmetic  an  improper  fraction  may  be  reduced  to  an  entire 
numler  plus  a  proper  fraction,  so  in  Algebra  an  improper  frac- 
tion may  be  reduced  to  an  entire  function  of  a  symbol  plus  a 
proper  fraction. 

EXERCISE^. 

Execute  the  following  divisions,  and  reduce  the  quotients 
to  the  form  {a)  when  there  is  any  remainder. 

1.  Divide  x^  —  2x  —  \  by  x  -\-\, 

2.  Divide  x?  -\- 2x^  —  %x  —  \  by  i?:;  —  1. 

3.  Divide  x^  —  Zx^  +  2a;  —  1  by  a;^  —  ir. 

2^:4  __  2^:3  _|_  ^2  _  ^  _  5 

4.  Keduce 

x^  —  x  —  \ 

5.  Divide  24^3  —  38«2  —  32a  +  50  by  2a  —  3. 

Ans.  Qaot.  =  12^2  —  a  — —\    Rem.  —  —-> 


64  ALGEBRAIC   OPERATIONS, 

6.  Divide  :r*  —  1  by  a:  —  1. 

When  terms  are  wanting  in  the  dividend,  they  may  be  considered  rs 
zero.     In  this  last  exercise,  the  terms  in  x^,  x-,  and  x  are  wanting.     But 
the  beginner  may  write  the  dividend  and  perform  the  operation  thus  : 
a^  +  Ox^  +  Ox^  +  Oaj  -  1   |  x-  1 

^  —   ^ aj3  +  a;2  +  a;  +  1 


x^ 

+ 

Ox' 

x^ 

- 

X' 

aj-^ 

+ 

Ox 

of' 

— 

X 

x-1 

Xj-1 

0      0 
The  operation  is  thus  assimilated  to  that  in  which  the  expression  is 
complete  ;  but  the  actual  writing  of  the  zero  terms  in  this  way  is  un- 
necessary, and  should  be  dispensed  with  as  soon  as  the  student  is  able 
to  do  it. 

7.  Divide  a^  —  2«  +  1  by  a  —  1. 

8.  Divide  x^  -{-  1  hj  x  -{-  1, 

9.  Divide  Sa^  +  125  by  2a  -+  5. 

10.  Divide*  ^^^  +  1  by  a  +  1. 

1 1.  Divide  a^  +  2^2  +  9  by  a^  -]- 2a  -^  3. 

12.  Divide  a^  —  1  hj  a^  -{-  2a^  +  2a  +  1. 

13.  Divide  x^  —  12a;4  +  36x^  —  32  by  x^  —  2. 

14.  Divide  {x^  —  2x-^  1)  {x^  —  12^'  —  16)  by  x"^  —  IG. 

For  some  purposes,  we  may  equally  well  perform  the  operation  by 
beginning  with  the  term  containing  the  lowest  power  of  the  quantity, 
or  not  containing  it  at  all.     Take,  for  instance,  Example  g  : 

125  +  Sa^    \  5  +  2a 

^^^  +  50a       25  -  10a  +  AcC' 
-60a 
-  50a  -  20a:' 

20a'^  +  8a3 
20a^  +  Sa^ 

15.  Divide  1  +  3:?;  +  3x^  +  x^  by  1  +  x. 

16.  Divide  1  —  4:X  -{-  4x^  —  x^  hj  1  —  x. 

17.  Divide  15  +  2«  ~  da^  +  a^  +  2a^  —  a^  by  5  +  4<^  —  a\ 

18.  Divide  1—f  by  1  -f-  2?/  -f  2^2  ^  y\ 

19.  Divide  64— 64:^:  +  16:z:2_8^_l_4^_^  by  —  4  +  2:?;  +  ^^. 

20.  Divide  64  —  16^^:^  j^  ofi  by  4  —  4a;  +  x^. 


DIVI8I0JS'.  65 

97.  Case  II.  When  there  are  several  algebraic  sym- 
hols  in  the  divisor  and  dividend. 

Let  us  suppose  the  dividend  and  divisor  arranged  accordiug 
to  powers  of  some  one  of  the  symbols,  which  we  may  suppose 
to  be  X,  as  in  §  76. 

Let  us  call  A  the  coefficient  of  the  highest  power  of  x  in 
the  dividend,  and  ^  the  term  independent  of  x^  so  that  the 
dividend  is  of  the  form 

Ax'^  +  (terms  with  lower  powers  of  x)  +  H. 

Let  us  call  a  the  coefficient  of  the  highest  power  of  x  in 
the  divisor,  and  h  the  term  of  the  divisor  independent  of  x,  so 
,that  the  divisor  is  of  the  form 

ax'^  +  (terms  with  lower  powers  of  x)  +  li. 

Then  we  have  the  following 

Theorem.  In  order  that  the  dividend  may  be  exact- 
ly divisible  by  the  divisor,  it  is  necessary : 

1.  That  the  term  containing  the  highest  power  of  x 
in  the  dividend  shall  be  exactly  divisible  by  the  cor- 
responding term  of  the  divisor. 

2.  That  the  term  independent  of  x  in  the  dividend 
shall  be  exactly  divisible  by  the  corresponding  term  of 
the  divisor. 

Reason,  The  reason  of  this  theorem  is  that  if  we  suppose 
the  quotient  also  arranged  according  to  the  powers  of  x,  then, 

1.  The  highest  term  of  the  dividend,  Ax'^,  will  be  given  by 
multiplying  the  highest  term  of  the  divisor,  aaP',  by  the  high- 
est term  of  the  quotient.     Hence  we  must  have, 

Ax^ 

Highest  term  of  quotient  =  — ^• 

2.  The  lowest  term  of  the  dividend  will  be  given  by  multi- 
plying the  lowest  term  of  the  dividend  by  the  lowest  term  of 
the  quotient.     Hence,  we  must  have, 

TT 

Lowest  term  of  quotient  =  -j-  • 

Eem.  1.  Since  we  may  arrange  the  dividend  and  divisor 
according  to  the  powers  of  any  one  of  the  symbols,  the  above 


66  ALGEBRAIC   OPERATIONS. 

theorem  must  be  true  whatever  symbol  we  take  in  the  place 
of  X. 

Eem.  2.  It  does  not  follow  that  when  the  conditions  of 
the  theorem  are  fulfilled,  the  division  can  always  be  performed. 
This  question  can  be  decided  only  by  trial. 

We  now  reach  the  following  rule : 

I.  Arrange  both  dividend  and  divisor  according  to 
the  ascending  or  descending  powers  of  some  coimnon 
symbol. 

II.  Form  the  first  term  of  the  quotient  by  dividing 
the  first  terjn  of  the  dividend  by  the  first  term  of  the 
divisor, 

III.  Multiply  the  whole  divisor  by  the  term  thus 
found,  and  subti^act  the  product  from  the  dividend. 

IV.  Treat  the  remainder  as  a  new  dividend  in  the 
same  way,  anfid  repeat  the  process  until  a  remainder  is 
found  which  is  not  divisible  by  the  quotient. 

Ex.  I.    Divide  x^  +  3ax^  +  Sa^x  +  a^  by  x  +  a. 

OPERATION. 

x^  +  dax^  +  3a^x  +  a^  \  x -{- a 

^+    ^^^  x^  +  '^ax  +  a^ 

2ax^  4-  da^x 
2ax^  +  2a^x 

d^x  +  a^ 
cfix  +  a^ 
"O        O" 
Ex.  2.    Divide  x^  —  ax^  -\-  a  (h -\- c)  x  —  ahc—bx^—cx^ + hex 


by  X  —  a. 

Arrang 
x^  —  (a-\-h-\-c)x^  -^  {al)-\-'bc-^ca)x  —  abc  \  x  —  a 


I 
Arranging  according  to  §  76,  we  have  the  dividend  as  follows :  | 


^— O^ x^—{b^c)x-\-ic 

—  {b-}-c)x^-{-  {ah-\-bc-\-ca)x 

—  {b-\-c)x^-\-  (ah  +  ac)  x 

hex  —  ahc 
hex  —  ahc 
~0         0 


FRACTIONS.  67 

EXERCISES, 

1.  Divide  the  dividend  of  Ex.  2  above  by  x  —  h. 

2.  Divide  the  dividend  of  Ex.  2  above  by  x  —  c. 

3.  Divide  a^  +  ¥—  (^  +  dabc  hj  a  +  b  —  c. 

4.  Divide  a^  -^¥  +  3ab  —  l  hj  a  +  b  —  1. 

5.  Divide  u^I^  +  2abx^  —  {a^  -f  I^)x^  by  ab  +  {a  —  b)  x. 

6.  Divide  (a^  —  bcf  -f  ^bV  by  o?  +  be, 

7.  Divide  (a  +  5  +  c)  (a2>  -{•  be  -\-  ea)  —  abe  hj  a  +  b. 

8.  Divide  {a  +  b  —  e){b  -\-  c —  a)  {e  -{- a —  b) 

by  a^  —  52  —  c2  H-  25c. 

9.  Divide  a^  +  5^  +  c^  —  3a5c  })j  a  -\-b  +  c. 

10.  Divide  a;^  -f-  4a^  by  a;^  —  2^a;  +  "Ha^ 

1 1.  Divide  a^{b  +  x) —  bf^ix  —  a)  +  {a  —  b)oo^  +  abx 

by  ic  +  a  +  i 

12.  Divide  01^  —  ax^  —  5^:^;  -\-  aW  by  (ic  —  t^)  (a;  +  b), 

13.  Divide  12a%9  —  14^5^  +  6^^^  —  a'  by  %a^7?  —  «3. 


■♦♦♦- 


CHAPTER     IV. 
OF    ALGEBRAIC     FRACTIONS. 

98.  Bef.  An  Algebraic  Fraction  is  the  expression 
of  an  indicated  quotient  when  the  divisor  will  not  ex- 
actly divide  the  dividend. 

Example.     The  quotient  of  ;?  -7-  ^  is  the  fraction  —  • 

Def. .  The  numerator  and  denominator  of  a  frac- 
tion  are  called  its  two  Terms. 

Transformation  of  Single  Fractions, 

99.  Reduction  to  Lowest  Terms,  If  the  two  terms 
of  a  fraction  are  multiplied  or  divided  Iby  the  same 
quantity,  the  value  of  the  fraction  will  not  be  altered. 


68  ALGEBRAIG   OPERATIONS, 


CLCC 

Example.     Consider  the  fraction  —    If  we  divide  both 

ay 

terms  by  a,  the  fraction  will  become  -• 

J  y 

ax X 

^'~  y 

Corollary,  If  the  numerator  and  denominator  con- 
tain common  factors,  they  may  be  cancelled. 

Def,  When  all  the  factors  common  to  the  two 
terms  of  a  fraction  are  cancelled,  the  fraction  is  said  to 
be  reduced  to  its  Lowest  Terms. 

To  reduce  a  fraction  to  its  lowest  terms,  factor  hoth 
terms,  ivhen  necessary,  and  cancel  all  the  common 
factors. 

Ex.  I.     — ^  =  — . 
acny'^       en 

The  factor  ay^  common  to  both  terms  is  cancelled. 
-^  .  aW       a^ 

The  factor  aW  common  to  both  terms  is  cancelled. 

Ex.  3.     Keduce  -1=-- 
aH 

Here  a^x  is  a  divisor  of  both  terms  of  the  fraction.     Di- 

1  o^x         1 

vidinff  by  it,  the  result  is  -^»    Hence  -^-  =  -* 
°    "^  (T  (vx        a^ 

-r,  mu  —  nu       (m  —  n)u       u 

Ex.  5.     =  ) ^  =  -. 

mx  —  nx       (m  -—  n)x       x 

EXERCISES. 

Reduce  the  following  fractions  to  their  lowest  terms  : 

aWp^  am 

I.     — -^—*  2.     • 

a^¥p  ahnx 

lOpqr^  12axy 


FRACTIONS. 


69 


5- 

7. 

9- 

II. 

IS- 
17. 


72(6g  — :r)(^-g) 

o^y  —  'by  _ 
ax  —  hx 


a"- 


W 


•a^  —  ^ah  +  l^ 

x^  +  y^ 
~a{x-\-'y) 
a^ 


U 


a^  -  V'^ 


f 


x^  —  y^ 
axm  —  axn 
hym  —  hyn 


10 
12 

14 
16 


20  {a  +  x){m  —  n) 
24  {a^—'ilax  +  x^){m—n) 
ahf  —  b^y^ 
ay  — by  ' 
a^  +  4:ax  +  4:X^ 

a^  —  4:X^ 
a^  +  8Z^3 
ay  +  2J?/  V 

mx  —  nx 
{a  -\-  b)  {m  —  n) 


100.  Rule  of  Sifpis  in  Fractions,  Since  a  fraction  is  an 
indicated  quotient,  the  rule  of  signs  corresponds  to  that  for 
division.  The  following  theorems  follow  from  the  laws  of 
multiplication  and  division : 

1.  If  the  terms  are  of  the  same  sign,  the  fraction  is 
positive ;  if  of  opposite  signs,  it  is  negative. 

2.  Changing  the  sign  of  either  term  changes  the 
sign  of  the  fraction. 

3.  Changing  the  signs  of  both  terms  leaves  the  frac- 
tion with,  its  original  sign. 

4.  The  sign  of  the  fraction  may  be  changed  by 
changing  the  sign  written  before  it. 

5.  To  these  may  be  added  the  general  principle  that 
an  even  number  of  changes  of  sign  restores  the  fraction 
to  its  original  sign. 

1'~~irb-       F"""~  ^=^' 

a  —  a —  a  a 

b         b  —  a  a  —  b  h  —  a 


Ex. 


Ex.2.     -%  = 


Ex.  3, 


m  —  n 


n  —  m 


m 


m  —  n 


70  ALGEBRAIC   OPERATIONS, 

EXERCISES. 

Express  the  following  fractions  in  four  different  ways  with 

respect  to  signs : 

X  —  y 

I.     -* 

a 

m 


X 

-y 

a 

a 

a 

-^  + 

c 

a 

+  m  — 

'X 

p-q 

m  —  n 
p -\- q  —  r  "a  —  m  +  x 

Write  the  following  fractions  so  that  the  symbols  x  and  y 
shall  be  positive  in  both  terms  : 

X  —  h  ^  m  —  X 

7.     H 8. 

c-y 

a  -\-  X  —  b 
9'      + 


I 

n  - 

-y 

a  — 

■  X 

J- 

■X 

a 

+  ^- 

—  X 

a  —  X  -\-  b 

X  —  a  -\-b 

II. ^ •  *«.  , 

b  —  X  a  —  b  -\-  y 

101.  When  the  numerator  is  a  product,  any  one  or 

more  of  its  factors  can  be  removed  from  the  numerator 

and  made  a  multiplier. 

^        abmx          ^    mx           7         ^  7  1 

Ex. =  ab ==  abm =:  abmx- 


p  -vq        p  +  q  p  -\-q  p  -\-  q 

EXERCISES. 

Express  the  following  fractions  in  as  many  forms  as  possi- 
ble with  respect  to  factors  : 

pax  ab  abc 

I.     -— •  2.     — .  3.     — -T« 

mn  c  a  -\-  0 

X^  —iP-  «4  _  J4  r^_  16^4 

^*      a  —b  ^'         X  '       X  +  2a 

103.  Reduction  to  Gimn  Denominator.  A  quan- 
tity may  be  expressed  as  a  fraction  with  any  required 
denominator,  i>,  by  supposing  it  to  have  the  denomi- 
nator 1,  and  then  multiplying  both  terms  by  D. 

For,  if  we  call  a  the  quantity,  we  have    ^  =  -  =  ^  • 


FRACTIONS.  71 

Ex.  If  we  wish  to  express  the  quantity  ab  as  a  fraction 
having  xy  for  its  denominator^  we  write 

ahxy 

If  the  quantity  is  fractional,  both  terms  of  the 
fraction  must  be  multiplied  by  that  factor  which  will 
produce  the  required  denominator. 

Ex.     To  express  t  with  the  denominator  nh^,  we  multiply 

both  members  by  n¥  -r-b  :=  nb^.     Thus, 

a  _  anV^ 

b  ~~  nb^ 

This  process  is  the  reverse  of  reduction  to  lowest  terms. 

EXERCISES. 

Express  the  quantity 

1.  a  with  the  denominator  b, 

2.  ax  "       "  " 

3.  ab  ''       " 

4.  —  "       ^^  " 


6 


-        1  an  u 

a  ■\-b 

H        ^  '^  y  u         cc  a 

^  —  y 

*       X    -\-l 

a  —  l 

Negative  Exponents. 
103.    By  the  principle  of  §  85,  we  have 

%  =  a^-K 
a^ 

If  we  have  hy  n,  the  exponent  of  the  second  member  of 

the  equation  will  be  negative,  and  the  first  member,  by  can-' 


n{x 

-y)' 

X. 

«2- 

■l\ 

a2- 

■f- 

x^  +  3a;  +  1. 

a*- 

•1. 

72  ALOEBRAIG   OPERATIONS. 

celling  n  factors  from  each  term  of  the  fraction,  will  become 
-^r— -•    Hence  -j~-  =  a^'K 

By  putting  for  shortness  k  —  n=:s,  the  equation  will  be 

1 

—  =  a'^ 
a' 

Hence, 

A  ne£ativ6  exponent  indicates  the  reciprocal  of  the 
corresponding  quantity  with  a  positive  exponent. 

If  in  the  formula  a^'^  =r  -^   we   suppose  h  :=  n,  it  will 

become  a^  =z  ~,  or  a^  -—1.     Hence,  because  a  may  be  any 

quantity  whatever, 

Any  quantity  with  the  exponent  0  is  equal  to  unity, 

Tliis  result  may  be  made  more  clear  by  suc- 
cessive divisions  of  a  power  of  a  by  a.  Every 
time  we  eftect  this  division,  we  diminish  the  ex- 
ponent by  1,  and  we  may  suppose  this  diminution 
to  continue  algebraically  to  negative  values  of 
the  exponent.  On  the  left-hand  side  of  the 
equations  in  the  margin,  the  division  is  effected 
symbolically  by  diminishing  the  exponents  ;  on 
the  right  the  result  is  written  out  in  the  usual 
way, 

EXERCISES. 

In  the  following  exercises,  write  the  quotients  which  are 
fractional  both  as  fractions  reduced  to  their  lowest  terms,  and 
as  entire  quantities  with  negative  exponents,  on  the  principle. 


d^ 

= 

aaa 

a' 

= 

aa 

a} 

= 

a 

a^ 

r= 

1 

a-i 

= 

1 

a 

a-2 

= 

1 
aa 

etc. 

etc. 

a 

ah-\ 

a:^ 

=:  a%-\ 

etc. 

Divide  . 

I.     x}  by  X. 

Ans.  X. 

2.  x  by  x^,      ' 

3.  _  2^^3  by  l\ 

Ajis.  -  or  x~'^. 

X 

-'^ov  -%a-W. 
a 

.  4.     ^aW'  by  —  "^a^. 

Ans. 

FRACTIONS.  73 

5.  —  ^a%  by  4:ad\  6.     \%a%^xy  by  ^abx. 

7.  14«^4^V  by  —  naWd^.         8,     Uapqxy  by  18a2»c. 

9.  —  dQ>a^p^x^y  by  —  2^a^xy. 

10.  486^2  (a;  —  2/)2  by  36  (iz:  —  y). 

11.  42^2^4'bj^Op-i^^y. 

\x  —  yf     ''        \x  —  yf 

12.  22  (a  —  V)  (m  —  ^)  by  15  [a  -\-h)(m  ■\-  n). 

13.  25  («2  —  Z>2)  ^^2  _  ^2)   by    15  (^  __  5)  (^^  ^  ^^), 

14.  (r^  -  1)  (a2  -  4^)  by  (ir2  -!)(«+  2^^). 

15.  i^:^  —  1  by  a^  +  1. 

16.  a^¥x^y^  by  a^b^x^y\ 

17.  mhi'^yH  by  mn^y^A 

18.  77i(m  +  l)  (m  +  2)  (m  +  3)  by  m  (m— 1)  (m— 2)  (m— 3). 

19.  a^  by  a**.  20.     a^^c^  by  ^J'^c'^. 

Dissection  of  Fractions. 

104.  If  the  numerator  is  a  polynomial,  each  of  its 
terms  may  be  divided  separately  Tby  the  denominator, 
and  the  several  fractions  connected  by  the  signs  +  or  — . 

The  principle  is  that  on  which  the  division  of  polynomials 
is  founded  (§  87).     The  general  form  is 

m  mm       m 

The  separate  fractions  may  then  be  reduced  to  their  lowest 
terms. 

Example.     Dissect  the  fraction 

d^aWx  —  l^amy  +  l^inz  —  I21^n^u 
IQabx 
The  general  form  (1)  gives  for  the  separate  fractions, 
d2aWx  __  18am^       Ubnz  _  12Pn^u 
IQabx        IQabx        16abx         16abx 
Eeducing  each  fraction  to  its  lowest  terms,  the  sum  becomes 
dmy      Ibnz       dbnho 
""  Sbx       16ax        4:ax 


74  ALGEBRAIC   OPERATIONS, 

EXKRCISES. 

Separate  into  sums  of  fractions, 
abc  +  hcd  +  cda  +  dab 

abed 
—  xyzu  +  x^yzu^  +  xyh'^u  —  x^yhV 


x^yh^v? 


(m  —  n){n  -^  q)  —  (m  -\-  n)(p  —  q) 

{m  —  n)  {])  —  q) 
(x  —  a){y  —  b)-j-  {x  ^y){a  —  b)  +  {x  —  b){y  —  a)  ^ 

x^  —  y'^ 
(a  4-  b)  {m  —  n)  —  {a  —  b){m  -{•  n) 


Agg^regation  of  Fractions. 

105.  When  several  fractions  have  equal  denomina- 
tors, their  sum  may  be  expressed  as  a  single  fraction 
by  aggregating  their  numerators  and  writing  the  com- 
mon denominator  under  them. 

m      m       m  m 

^  a  —  b      b  —  c    ,   c  —  a 

Ex.  2. 

^  —  y    y  —  ^    ^  — y 

_  a  —  b      c—b       c  —  a_2c  —  2b_2{c—'b) 
~  X  —  y      X  —  y      X  —  y  ~    x  —  y   ~~     x  —  y 
Hem.    This  process  is  the  reverse  of  that  of  dissecting  a  fraction, 

EXERCISES. 

Aggregate 

a        ab       abc  a  b 


abc      abc      abc  '     {a  —  by       {a  —  b)^ 

^  —  a    ,  y—  b      a-{-  b  x  —  y 
a^x           a^x           a^x  a^x 

c  d 


+ 


a—b      b — a      a  —  b      b  —  a 

a  —  b        a  —  c        c  —  b        c  A-  a 

5. 1 ±- — 

m  —  n      m  —  n      n  —  7n      n  —  m 


FRACTIONS.  75 

106.  When  all  the  fractions  have  not  the  same  denomina- 
tor, they  must  be  reduced  to  a  common  denominator  by  the 
process  of  §  102. 

Any  common  multiple  of  the  denominators  may  be  taken 
as  the  common  denominator,  but  the  least  common  multiple  is 
the  simplest. 

To    KEDUCE    TO     A     COMMOi^     DENOMINATOR.      CJlOOSe  a 

common  multiple  of  the  denominators. 

Multiply  both  terms  of  each  fraction  by  the  multi- 
plier necessary  to  change  its  denominator  to  the  chosen 
multiple. 

Note  1.  The  required  multipliers  will  be  the  quotients  of 
the  chosen  multiple  by  the  denominator  of  each  separate 
fraction. 

ISToTE  2.  When  the  denominators  have  no  common  fac- 
tors, the  multiplier  for  each  fraction  will  be  the  product  of  the 
denominators  of  all  the  other  fractions. 

Note  3.  An  entire  quantity  must  be  regarded  as  having 
the  denominator  1.     (§  102.) 

EXAMPLES. 

I.     Aggregate  the  sum 

1        1^  _  J^       J\_ 

a      ab      cibc      abed 
in  a  single  fraction. 

The  least  common  multiple  of  the  denominators  is  abccL 
The  separate  multipliers  necessary  to  reduce  to  this  com- 
mon denominator  are 

abed,     bed,     cd,    d,    1. 

The  fractions  reduced  to  the  common  denominator  abed  are 

abed      —  bed      -\-  ed      —  d      +1 
abed^      abed '     abed '    abed^    abed 

^,  .         aled  —  bed  -\-  ed  —  d  -\-  1 

The  sum  is ^r-i » 

abed 

By  dissecting  this  fraction  as  in  §  104,  it  may  be  reduced 
to  its  original  form. 


76  ALGEBRAIC   OPERATIONS. 

2.  Eeduce  the  sum 

1       ah      c 
a      T)      c'~'^ 
to  a  single  fraction. 

The  multipliers  are.  by  Note  2,  bed,  acd,  aid,  ale. 

Using  these  multipliers,  the  fractions  become 
led       —  a^cd      a¥d       —  al(^ 
abcd^       alcd  '     alcd^      alcd  ' 

from  which  the  required  sum  is  readily  formed. 

3.  Eeduce  the  sum 

1  X       ^       01? 

"^  x  —  1  "^  2^  +  1  ■^"  ^-^* 
The  least  common  multiple  of  the  denominators  is  x^  —  1. 
The  multipliers  are,  by  Note  1, 

x^  —  1,    X  -\-l,     X  —  1,     1. 
The  sum  of  the  fractions  is  found  to  be 

x'^  —  1  -{-  X  -{■  1  -{-  x^  —  X  +  x^  __     3x^^ 
x^  —  1  ~~  x^  —  1 

EXERCISES. 

Eeduce  to  a  single  fraction  the  sums, 


3 

5 
7 
9 
10. 

II 


11  11 

4.  -, — z  + 


1  —  i?;l  +  a;  1  —X      1  +  X 

ax  x^  .  a  I 

6. 


a  +  X      a  +  X  'a  —  I      a  -\-  I 

a  X  ^       2x  —  6  6  3 


x(a  —  x)       a{a  —  x)        '     4:X^  —  1       2a;  —  1       x 

1     I     gy         1 


X  -}-  y       x^  —  y^      X  —  y 
1  1 


+  tT— -  + 


a— I      I  —  c      c  —  a 
fi.  a 

+  - — -;  12 


a  ■\-l      a— I 

X  ■\-  y  '   x  —  y  *"'     a—'l^  a-Vl 


FRACTIONS.  77 

«2  -f.  J2  I  a 

1 1_ 1 

^4-     ^{x-1)       ^{x  -{-1)      x^ 
_a L ^\ 

^^'     a  —  b       \         a  —  b/' 

m  -\-  n      ^  —  y  y  m  -\-  y 

m  —  n      X  -\-  y  m^      m  {m  —  y) 

^      ^  a  x^ 


20. 


\a-b^  b-al 


a  —  X      c^  —  x^ 
a  —  b       1)  —  c      c  —  a       {a  —  b)  (b  ■—  c)  {c  —  a) 
'^'     a~+~b  "^  b~+~c  "*"  c  -{-  a  "^  {a  +  b)  {b  +  c)  {c~^aj' 
a 
b 
m  —  (x  —  a)       m  —  (x  -}-  a) 

21.       ^ ^^ -» 

X  -{-  y  x  —  y 

c    ^    a        h 

22.  -^  +  —  -^ 

ab      be      ac 

a b c      

^3-     (^  _  ^)  (^  _  c)  +  (^  _  a)  {b-c)'^  {c-a)  (c-b) 

X  -\-l       X  —  1        , 
24. +  ^x. 

X  —  1         X  -\-l 

ab  a^         a  {a^  +  P) 

26.        1 


27.        1  — 


X  -\-  a      X  —  a 
x^  —  '^xy  +  y^ 


x^  +  y'^ 


28      1  —  ?— ±-^^-Zl^l 

1^ 1_  Jl 

^9-     (^  _!_  j^y  +  («  -  Z>)2  +  a2  -  6« 

a2  _  <^ab  +  ^>2 
3^-     ^  + 4^1 


78  ALGEBRAIC   OPERATIONS. 


Factoring  Fractions. 

107.  If  several  terms  of  the  numerator  contain  a 
common  factor,  the  coefficients  of  this  factor  may  be 
added,  and  their  aggregate  multiplied  by  the  factor  for 
a  new  form  of  the  numerator. 

EXAMPLES. 

ax  — -  hx  +  ex  +  dx  _  {a  —  b  +  c  -\-  d)  x 
m  ~  m 

=^{a-h  +  c  +  d)^^.    (§101.) 

dbx  +  hex  +  acy  —  dby  __  {ab  -\-  he)  x       {ac  —  ah)  y 
ahn  ~~    ahn  ahn 

=  (a  +  c)^^  +  ic-l)l- 

EXERCISES. 

Eeduce 

ahy  —  hey  —  acy  mnu  +  7npu  +  pnu 


3 

4 

5 
6 
8, 
9 

10 


ahe  '  mn 

ahq  +  hcq  +  ahr  4-  her 

ahe 
ax  —  hy  —  dhx  —  4:ay 

2ma 
4:7nx  -i-  2y  —  dax  —  6ex  -f  ay 

xyz 
a^  +  ^a^h  +  al^  a^x  —  ^ahe  —  (3^/  —  4g)  o? 

xy  '  '*  i?  +  S' 

a;2y  —  [4a;  -f  a;  {2h  --  4g)  +  3^cg] 

a-^h 
ax^  —  4ca;  —  3  \mx  -\-  m  {a  —  x)  —  am] 
2a  —  dh 

4:aVx  — -  2eV^  +  2hVx  —  2  iif)in\/x  —  4:Vx). 
'~'  da  —  ^b 


FRACTIONS.  70 


Multiplication  and  Division  of  Fractions. 

108.  Fundamental  Theorems  in  the  Multiplication 
and  Division  of  Fractions : 

Theorem  L  A  fraction  may  be  multiplied  by  any 
quantity  by  either  multiplying  its  numerator  or  dividing 
its  denominator  by  that  quantity. 

Cor,  1.  A  fraction  may  be  multiplied  by  its  denominator 
by  simply  cancelling  it. 

Cor.  2.    If  the  denominator  of  the  fraction  is  a  factor  in 

the  multiplier,  cancel  the  denominator  to  multiply  by  this 

factor,  and  then  multiply  the  numerator  by  the  other  factors. 

m 
Ex.        — -; TT  X  a^  {x^  —  IP)  ==  mn  {x  +  h), 

Ct  \pC  —  0) 

because  the  multiplier  «2  (^2  _  ^2^  =  a{x  —  b)a  {x  +  b). 

Theorem  11.  A  fraction  may  be  divided  by  either 
dividing  its  numerator  or  multiplying  its  denominator. 

Theorem  IIL  To  multiply  by  a  fraction,  the  multi- 
plicand must  be  multiplied  by  the  numerator  of  the 
fraction,  and  this  product  must  be  divided  by  its  de- 
nominator. 

Let  us  multiply  t  by  — 

We  multiply  by  m  by  multiplying  the  numerator  (Th.  I), 
and  we  divide  by  n  by  multiplying  the  denominator  (Th.  II). 

Hence  the  product  is         y— • 

That  is,  tJxe  product  of  the  numerators  is  the  numer- 
ator of  the  required  fraction,  and  the  product  of  the 
denominators  is  its  denominator, 

EXERCISES. 

Multiply 

ab  ■\-  y  ,  ah  ^     X 

I. m  X  —  a,  2,     —  by  -• 

X  —'  a     '^  X     ^  a 

ab    .  «^     u      9        9 

3.    -zTx  ^y  ^y-      .  4.     --::z^  by  x^  -  a\ 


80  ALGEBBAIO   OPERATIONS. 

^       xhj      -^     ^  x^     -^  x  —  m 

7.     ^Zl_*by^  +  i.  8.     «  +  !^by«  +  ^. 

iz; ,  v—ab  m  -\-  n  .     n  —  m 

g.     ab by  ay  + 10. by  , 

^  y  X  7n  —  n    ^  7n  -\-  n 

Ti^  ii-  1  Ix  ^    a      h      X 

11.  Multiply  «+-  by  3  +  -  +  -- 

^  ^        /  mn   \  I  mn   \ 

12.  Eeduce  [m  -\ ]\m ■ — )• 

v  711  —  nJ  \  m  -\-  n1 

13.  Eeduce  \a  -  "f)  (^  "  y)* 

14.  Multiply  ^  -  ^  by  ^. 

Ans*  — 


15. 

Divide  —  by  p. 

16. 

Divide  7  by  a  +  J. 

a  —  0 

17. 

Divide  — — ■  by  x  —  1. 
iC  +  1 

18. 

Divide   ^^  +  J  by  1  +  xK 

19. 

^.  .,     —  2a  —  3m  ,      ,^ 
DiYide  — T-~  by  J'*  —  a^. 

109.  Reciprocal  of  a  Fraction.    The  reciprocal  of 
a  fraction  is  formed  by  simply  inverting  its  terms. 

Eor,  let  T  be  the  fraction.    By  definition,  its  reciprocal 

will  be 

1^ 

a 

b 

Multiplying  both  terms  by  b^  the  numerator  will  be  b  and 

the  denominator  z  x  b,  that  is,  a. 

Hence  the  reciprocal  required  will  be  -,  or,  in  algebraic 
language, 


I 


FRACTIONS.  81 

a  ^  a 
b 
110.  Def.     A  Complex  Fraction  is  one  of  which 
either  of  the  terms  is  itself  fractional. 

a 
b 


Example. 


m  A — 

y 


a  OR 

is  a  complex  fraction,  of  which  ^  is  the  numerator,  and  m  +  - 

the  denominator. 

The  terms  of  the  lesser  fractions  which  enter  into  the 
numerator  and  denominator  of  the  main  fraction  may 
be  called  Minor  Terms. 

Thus,  b  and  y  are  minor  denominators,  and  a  and  x  are 
minor  numerators. 

To  reduce  a  complex  fraction  to  a  simple  one,  multi- 
ply both  terms  hy  a  multiple  of  the  minor  denominators. 

am 

Example.    Eeduce 


y2 


Multiplying  both  terms  by  xy^,  the  result  will  be 
amx 


bxy  +  Jiy^* 
which  is  a  simple  fraction. 

EXERCISES. 

Reduce  to  simple  fractions  : 

X  b 

a  ^  - 

X 


I. ^*  2. 


1 

-^i 

1 

a 

X 
—  X 

a 

-{-X 

b 

a 

X 


ab 
mil 
a  -\-  X  ^'     ~bd' 

a-^x  km 

6 


82  ALGEBBAIG   OPERATIONS, 


1  +  ^"' 

n  +  1 

am  -\ 

h 

an 

n 

2ab 

a^  +  l  +  2 

a 

a  +  2b      a 

a  -{-  2h          a 

b           a  -\-b 

8. 


II. 12. 


^3.      ^   ,    o. 1 14. 


l-\-x 

1  — 

X 

1—x 

+ 

1  + 

X 

1  -{-  X 

1  — 

X 

1—x 

1  + 

X 

2x- 

3 

'  y 

a  -{-  b 

— 

x 

1 

+ 

a 

I  +  a 

1  — 

a 

1 

a 

1-a 

r+ 

a 

1 

a 

a      1 

1 

b       b 

+ 

a 

x-y 

+ 

y  + 

X 

x^y 

f- 

■  x^ 

x-\-y 

x^- 

f 

X  —  y       x^  —  y^ 

Division  of  one  Fraction  by  Another. 

a        711/ 
111.  Let  us  divide  t  by  —    The  result  will  be  expressed 

by  the  complex  fraction 

a 
b 
m 
n 
Reducing  this  fraction  by  the  rule  of  §  110,  it  becomes 

an 
bm^ 

which  is  equal  to  t  x  —  •        That  is, 

b       m  ' 

To  divide  hy  a  fraction,  we  have  only  to  Tnultiply  by 
its  reciprocal. 


Divide 


FRACTIONS.  83 


EXERCISES. 


ab     ,     a  X  -i-  1  ,     2x 


X      ^    X  a^  —  ¥       ,     a^-[-db 


5. 


r  +  1  ,       a;  4-  1  ^      «       m  ,      h       n 

—— T  by  -^ T-  6.     y  H by 

X---1    "^  x^  —  1  0       n     ''  a      m 


a       h      c  ,      m       n       p 

7.     -4--4.-by h--i--- 

*      x      y      z    '    x       y       z 


a—-  b      a  -\-  b    ^  a—b      a  ■\- b 


Reciprocal   Relations    of  Multiplication   and 
Division. 

113.  The  fundamental  principles  of  the  operations  upon 
fractions  are  included  in  the  following  summary,  the  under- 
standing of  which  will  afford  the  student  a  test  of  his  grasp  of 
the  subject. 

1.  The  reciprocal  of  the  reciprocal  of  a  number  is 
equal  to  the  number  itself.    In  the  language  of  Algebra, 

1 

T  =  ^- 
a 

2.  The  reciprocal  of  a  monomial  may  be  expressed 
by  changing  the  algebraic  sign  of  its  exponent. 

3.  To  multiply  by  a  number  is  equivalent  to  dividing 
by  its  reciprocal,  and  vice  versa.     That  is, 

jy  X  a    or    —  =  aN, 
a 

and  vice  versa,  iv  x  -  =  — 

a        a 


84  ALGEBRAIC    0PEBAT10N8. 

4.  When  the  mimerator  or  denominator  of  a  fraction 
is  a  product  of  several  factors,  any  of  these  factors  may 
be  transferred  from  one  term  of  the  fraction  to  the  other 
by  changing  it  to  its  reciprocal.     That  is. 


,  7  -  abc 

abc DC  p 

par  ~~  1  ~~     qr 

^^  -  par  ^ 


etc. 


^  abc  Ic  p~^dbc 

Or,  =  -— j = ,    etc. 

pqr       a  ^pqr  qr 

5.  Mxdtiplicaticm  by  a  factor 

greater  than  nnity  increases^ 
less  than  unity  diminisJies. 
Bivision  by  a  divisor 

greater  than  unity  diminishes^ 
less  than  unity  increases. 

6.  («)  When  a  factor  becomes  zero,  the  product  also 
becomes  zero. 

{^)  When  a  denominator  becomes  zero,  the  quotient 
becomes  infinite.    That  is, 

Oxa==:«xO  =  0. 

-  =:  infinity. 

Note.  The  following  way  of  expressing  what  is  meant  by 
this  last  statement  is  less  simple,  but  is  logically  more  correct : 

If  a  fraction  has  a  fixed  numerator,  no  matter  how 
small,  we  can  make  the  denominator  so  much  smaller 
that  the  fraction  shall  be  greater  than  any  quantity  we 
choose  to  assign. 

EXERCISE. 

If  the  numerator  of  a  fraction  is  2,  how  small  must  the 
denominator  be  in  order  that  the  fraction  may  exceed  one 
thousand?  That  it  may  exceed  one  million?  That  it  may 
exceed  one  thousand  millions  ? 


BOOK    III. 
OF    E  QUA  TIONS. 

CHAPTER    I. 
THE     REDUCTION     OF     EQUATIONS. 


Definitions. 

113.  Def.  An  Equation  is  a  statement,  in  the  lan- 
guage of  Algebra,  that  two  expressions  are  equal. 

114.  Def.  The  two  equal  expressions  are  called 
Members  of  the  equation. 

115.  Def.  An  Identical  Equation  is  one  which  is 
true  for  all  values  of  the  algebraic  symbols  which  enter 
into  it,  or  which  has  numbers  only  for  its  members. 

Examples.    The  equations 

14  +  9  r=  29  —  6, 
5  _l_  13  _  3  X  4  —  6  =  0, 
which  contain  no  algebraic  symbols,  arc  identical  equations. 

So  also  are  the  equations 

X  ■=  X, 

X  —  X  ^=1   0, 

(x  -{-  a)  {x  —  a)  —  x^  —  a^, 

because  they  are  necessarily  true,  whatever  values  we  assign  to 
X,  a,  and  y. 

Eem.  All  the  equations  used  in  the  preceding  two  books 
to  express  the  relations  of  algebraic  quantities  are  identical 
ones,  because  they  are  true  for  all  values  of  these  quantities. 


86  EQUATIONS. 

116.  Def,  An  Equation  of  Condition  is  one  whicli 
can  Ibe  trne  only  when  the  algebraic  symbols  are  equal 
to  certain  quantities,  or  have  certain  relations  among 
themselves. 

Examples.    The  equation 

:c  +  6  =  22 

can  be  true  only  when  x  is  equal  to  16,  and  is  therefore  an 
equation  of  condition. 
The  equation 

X  -\-  h  =z  a 

can  be  true  only  when  x  is  equal  to  the  difference  of  the  two 
quantities  a  and  h 

Rem.  In  an  equation  of  condition,  some  of  the  quantities 
may  be  supposed  to  be  known  and  others  to  be  unknown. 

117.  Def.  To  Solve  an  equation  means  to  find 
some  number  or  algebraic  expression  which,  being  sub- 
stituted for  the  unknown  quantity,  will  render  the 
equation  identically  true. 

This  value  of  the  unknown  quantity  is  called  a  Root 
of  the  equation. 

EXAMPLES. 

1.  The  number  3  is  a  root  of  the  equation 

2a;2  _  18  m  0, 

because  when  we  put  3  in  place  of  x,  the  equation  is  satisfied 
identically. 

2.  The  expression  is  a  root  of  the  equation 

"zcx  —  4:a  +  2b  =  0, 

when  x  is  the  unknown  quantity,  because  when  we  substitute 
this  expression  in  place  of  x,  we  have 

2c(^^^)^^a  +  2b  =  0,  ^ 

or  ^a  —  2b  —  4ca-{-2b  =  0, 

which  is  identically  true. 


AXI0M8.  87 

Eem.  It  is  common  in  Elementary  Algebra  to  represent 
unknown  quantities  by  the  last  letters  of  the  alphabet,  and 
quantities  supposed  to  be  known  by  the  first  letters.  But  this 
is  not  at  all  necessary,  and  the  student  should  accustom  him- 
self to  regard  any  one  symbol  as  an  unknown  quantity. 

Axioms. 

118.  Def.  An  Axiom  is  a  proposition  wMch  is 
taken  for  granted,  without  proof. 

Equations  are  solved  by  operations  founded  upon  the  fol- 
lowing axioms,  which  are  self-evident,  and  so  need  no  proof. 

Ax.  I.  If  equal  quantities  be  added  to  the  two 
members  of  an  equation,  the  members  will  still  be  equal. 

Ax.  II.  If  equal  quantities  be  subtracted  from  the 
two  members  of  an  equation,  they  will  still  be  equal. 

Ax.  III.  If  the  two  members  be  multiplied  by  equal 
factors,  they  will  still  be  equal. 

Ax.  IV.  If  the  two  members  be  divided  by  equal 
divisors  (the  divisors  being  different  from  zero),  they 
will  still  be  equal. 

Ax.  V.    Similar  roots  of  the  two  members  are  equal. 

These  axioms  may  be  summed  up  in  the  single  one. 
Similar  operations  upon  equal  quantities  give  equal 
results, 

119.  An  algebraic  equation  is  solved  by  performing 
such  similar  operations  upon  its  two  members  that  the 
unknown  quantity  shall  finally  stand  alone  as  one 
member  of  an  equation. 

Operations  of  Addition  and  Subtraction— Trans- 
posing Terms. 

130.  Theorem.  Any  term  may  be  transposed  from 
one  member  of  an  equation  to  the  other  member,  if  its 
sign  be  changed. 


88  EQUATIONS. 

Proof.     Let  us  put,  in  accordance  with  §  41,  2d  Prin., 
t,  any  term  of  either  member  of  the  equation. 
a,  all  the  other  terms  of  the  same  member. 
h,  the  opposite  member. 
The  equation  is  then 

Now  subtract  t  from  both  sides  (Axiom  II), 

or  by  reduction,  a  ^=  h  —  t. 

This  equation  is  the  same  as  the  one  from  which  we  started, 
except  that  t  has  been  transposed  to  the  second  member,  with 
ifcs  sign  changed  irom  +  to  — . 

If  the  equation  is 

b  —  t  =^  a, 

we  may  add  t  to  both  members,  which  would  give 

b  =  a  +  L 

NUMERICAL      EXAMPLE. 

The  learner  will  test  each  side  of  the  following  equations  : 

194.3_94-4  rr:  7  +  10. 

Transposing  4,  194-3  — 9  =  7  +  10— 4. 

9,  19  +  3  =  7  +  10—4  +  9. 

19,  3  =  7  +  10-4  +  9-19. 

"3,  0  =  7  +  10—4  +  9-19—3. 

131.  Rem.  All  the  terms  of  either  member  of  an 
equation  may  be  transposed  to  the  other  member, 
leaving  only  0  on  one  side. 

Example.    If  in  the  equation 

b  =  a  +  t, 
we  transpose  b,  we  have    0  =  a  +  t  —  b. 

By  transposing  a  and  t,  we  have 

b-^a^t  =  0. 

123.  Changing  Signs  of  Members.  If  we  change  the  signs 
of  all  the  terms  in  hoth  memhers  of  an  equation,  it  will  still 
be  true.    The   result  will  be  the  same  as  multiplying  both 


REDUCTION.  89 

members  by  —  1,  or  transposing  all  the  terms  of  each  member 
to  the  other  side,  and  then  exchanging  the  terms. 

Example.     The  equation 

17  +  8  =  11  +  14 
may  be  transformed  into     0  =  11  +  14  —  17  —  8, 

or,  0  =  —  11  —  14  +  17  -f  8, 

or,     — 17  -  8  =z  —  11  _  14. 

Operation  of  Multiplication. 

133.  Clearing  of  Fractions,  The  operation  of  multipli- 
cation is  usually  performed  upon  the  two  sides  of  an  equation, 
in  order  to  clear  the  equation  of  fractions. 

To  clear  an  equation  of  fractions : 

First  Method.  Multiply  its  members  by  the  least 
common  multiple  of  all  its  denominators. 

Secoi^d  Method.  Multiply  its  members  by  each  of 
the  denominators  in  succession, 

Eem.  1.  Sometimes  the  one  and  sometimes  the  other  of 
these  methods  is  the  more  convenient. 

Rem.  2.  The  operation  of  clearing  of  fractions  is  similar 
to  that  of  reducing  fractions  to  a  common  denominator. 

Example  of  First  Method.  Clear  from  fractions  the 
equation 

Here  24  is  the  least  common  multiple  of  the  denominators. 
Multiplying  each  term  by  it,  we  have, 

Q,x-\-Ax  +  ^x  =  624, 
or  13:2;  r=  624. 

Example  of  Secokd  Method.     Clear  the  equation 


X  —  a      X  '\-  a      X 
Multiplying  by  x  —  a,  we  find 

ax  —  a^      ex  —  ca 

a  A 1 =  0. 

X  ■\-  a  X 


90  EQUATIONS. 

Multiplying  by  rr  -f-  «, 

ax  -\-  a^  -{-  ax  —  a^  -\ =  0. 

X 

Keducing  and  multiplying  by  Xy 

2ax^  +  cx^  —  ca^  =  0. 

EXERCISES. 

Clear  the  following  equations  of  fractions  : 

/iX  rt  r\  XX  ^_ 

--6  =  0.  ..     5-7  =  ^^- 

^^^_K  X      x^  _  b 

2+3~i-^-  4-     «  +  ««-«■ 

1  .y>l - J-.       6   ?.^-? 

ab^  a^  b~  aW  "     3  "^  4  ~  o' 


II 


13 


cc  —  a 

X  -\-  a 

X  ■\-  a 
X  —  a 

x^  4-  2ax 
X  —  a 

^_y 
y     ^ 

a 

X 

.    y   - 

10. 


14. 


a  —  ^       h  —  a  "" 
Here  the  second  term  is  the  same  as 

X  -{-  a       X  —h 


X  ^  a 

^  x+  b 

x  —  2 

x-^2 

x  —  6 

~'  X  +  b 

X  —  a 
X  -f  a 

X  -\-  a 
X  —  a 

-y 

a 


a  —  X       X  —  a 
Reduction  to  the  Normal  Form. 


( 


124.  Def,  An  equation  is  in  its  Normal  Form 
when  its  terms  are  reduced  and  arranged  according  to 
the  powers  of  the  unknown  quantity. 

In  the  normal  form  one  member  of  the  equation  is  expressed 
as  an  entire  function  of  the  unknown  quantity,  and  the  other 
is  zero.     (Compare  §§  50,  ?6.) 

To  reduce  an  equation  to  the  normal  form : 
I.  Transpose  all  the  terms  to  one  mewiber  of  the  equa- 
tion,  so  as  to  leave  0  as  the  other  mernber. 


REDUCTION,  9] 

II.  Clear  the  equation  of  fractions. 

III.  Clear  the  equation  of  -parentheses  hy  performing 
all  the  operations  indicated. 

lY.  Collect  each  set  of  tains  containing  lihe  powers 
of  the  unhnown  quantity  into  a  single  one. 

V.  Divide  hy  any  common  factor  which  does  not  con- 
tain the  unhnown  quantity. 

Eem.  This  order  of  operations  may  be  deviated  from 
according  to  circumstances.  After  a  little  practice,  the  student 
may  take  the  shortest  way  of  reaching  the  result,  without  re- 
spect to  rules. 

EXAMPLES. 

1.  Eeduce  to  the  normal  form 

(a;  —  2)  {x  —  3)  _  (x  +  2)  (a:  +  4) 
x  —  h  ~^  i?;  +  5 

1    Clearing  of  fractions, 

{x  +  5)  {x  —  2)  {x  —  Z)^(x  —  5)  (x  +  2)  {x  +  4). 

2.  Performing  the  indicated  operations, 

^  _  19^'  _{_  30  =  ^3  _|_  ^2  _  222;  —  40. 

3.  Transposing  all  the  terms  to  the  second  member  and 
reducing, 

^  z=.x^  —  ?>x—  70, 

which  is  the  normal  form  of  the  equation. 

Kem.  Had  we  transposed  the  terms  of  the  second  member 
to  the  first  one,  the  result  would  have  been 

~  a;2  +  3:^;  +  '^0  =  0. 

Either  form  of  the  equation  is  correct,  but,  for  the  sake  of 
uniformity,  it  is  customary  to  transpose  the  terms  so  that  the 
coefiicient  of  the  highest  power  of  x  shall  be  positive.  If  it 
comes  out  negative,  it  is  only  necessary  to  change  the  signs  of 
all  the  terms  of  the  equation. 

Ex.  2.     Reduce  to  the  normal  form, 

hmx^         2ax  ^mx^         ^ 

-„  z=^  2mx  —  D«. 

X  —  a       X  -{-  a       X?  —  ci'^ 


92  EQUATIONS. 

1.  Transposing  to  the  first  member, 

^mx^         2ax  3ma^ 

. __  2mx  -i-  6a  ~  0. 

X  —  a       X  -\-  a       x"  —  a'^ 

2.  To  clear  of  fractions,  we  notice  that  the  least  common 
multiple  of  the  denominators  is  x'^  —  a\  Multiplying  each 
term  by  this  factor,  we  have, 

6mx^{x+a)—2ax{x—a)—3mx^—2?nx{x^—a^)  +  6a(x^—a^)  =  0. 

3.  Performing  the  indicated  operations, 
6ma^+6amx^—2ax^+2a^x—dma^—2mx^-\-2a^mx+6ax^—6a^=0. 

4.  Collecting  like  powers  of  x,  as  in  §  76, 

{da  +  6am)  x^  +  {2a^  +  2a^m)  x  —  6a^  =  0, 

6,  Every  terra  of  the  equation  contains  the  factor  a.  By 
Axiom  IV,  §  118,  if  both  members  of  the  equation  be  divided 
by  a,  the  equation  will  still  be  true.  The  second  member 
being  zero,  will  remain  zero  when  divided  by  a.  Dividing 
both  members,  we  have 

(3  +  6771)  x^  +  2a{l+m)x^  6a^  =  0, 

which  is  the  normal  form. 

EXERCISES. 

Eeduce  the  following  equations  to  the  normal  form,  x,  y, 
or  z  being  the  unknown  quantity : 

dy^  -\-2y  __y  -—1  x  —  a__^x-{-a 

7        "~"2  '     X  -j-  a  ~~      X 


3- 


a;  —  7  2x-{-6 


2x  -h  10       4a;  —  2 


a^  ^  Za^x  +  2^3        ^      ^  n^  —  6ax^ 

2x  -\-  a  2x  —  a 

6.         '^-^-  +  -±-  =  0. 
a-\-bb-{'Z^a  +  z 

7?  z^  ah 


a  —z      a^  —  x^       ^  —  c^ 


REDUCTION.  93 


7  +  ^^S-^+i3=0. 

y     y^     y^ 

a  a?  ^^      _  1 

X  —  a      x^  —  a^      x^  —  a^ 

I)^  ¥  h^ 


+ 


c  —  z      c^  —  z^      d^  —  z^       c^  — 


Z"" 


a               b  mm 

II. = 12. = -. 

1       X  —  a  n            .1 

0 nx X  -^ — 

X  X                    tX/ 


13.      T  + 


1    '  1   ^  a^ 


a a' 


-2 


X  x 


14. 


15. 


16. 


_3z__  _     6z^     _  1 
I       ^        3  "  z' 

ax  hx 


1+   ' 


X  -{-  a  X  —  a 

a  h 


X      a  —  x  a 


b 

""       i 

a 

X 

X 

Degree  of  Equations. 

135.  Def.  An  equation  is  said  to  be  of  the  n^^  de- 
gree when  n  is  the  highest  power  of  the  unknown 
quantity  which  appears  in  the  equation  after  it  is  re- 
duced to  the  normal  form. 

EXAMPLES. 

The  equation       Ax  -{-  B  ^  0  is  of  the  first  degree. 

Ax'^  +  B  =z  0  "  "        second    " 

Ax^  +  Bx-\-  C  =::  0  "  "        third 

etc.  etc. 

An  equation  of  the  second  degree  is  also  called  a 
Quadratic  Equation. 


94  EQUATIONS  OF  THE  FIRST  DEGREE, 

An  equation  of  the  third  degree  is  also  called  a 
Oubic  Equation. 

Example.     The  equation 

ax^  +  hx^y^  -\-  y^  -\-  ah  z:^  0 
is  a  quadratic  equation  in  x,  because  x^  is  of  tlie  highest  power 
of  x  which  enters  into  it. 

It  is  a  cubic  equation  in  y. 

It  is  of  the  first  degree  in  z. 


CHAPTER    II. 

EQUATIONS    OF    THE    FIRST    DEGREE    WITH    ONE 
UNKNOWN    QUANTITY. 

136.  Eemark.  By  thQ  preceding  definition  of  the  degree 
of  an  equation,  it  will  be  seen  that  an  equation  of  the  first 
degree,  with  x  as  the  quantity  supposed  to  be  unknown,  is  one 
which  can  be  reduced  to  the  form 

Ax-\-  B  =  0,  {a) 

A  and  B  being  any  numbers  or  algebraic  expressions  which 
do  not  contain  x. 

Such  an  equation  is  frequently  called  a  Simple  Equation. 

Solution  of  Equations  of  the  First  Degree. 

137.  If,  in  the  above  equation,  we  transpose  the  term  B 
to  the  second  member,  we  have 

Ax  =  ^B. 

If  we  divide  both  members  by  A  (§  118,  Ax.  IV),  we  have^ 

B 

Here  we  have  attained  our  object  of  so  transforming  the 
equation  that  one  member  shall  consist  of  x  alone,  and  the 
other  member  shall  not  contain  x. 


ONE    UNKNOWN   QUANTITY.  95 

To  prove  that j  is  the  required  value  of  x,  we  sabsti- 

tute  it  for  x  in  the  equation  {a).     The  equation  then  becomes, 

or,  by  reducing,  —  ^  +  ^  =  0, 

an  equation  which  is  identically  true.  Therefore,  —  -^  is 
the  required  root  of  the  equation  (a),     (§  117,  Def.) 

138,  In  an  equation  of  the  first  degree,  it  will  be  unneces- 
sary to  reduce  the  equation  entirely  to  the  normal  form  by 
transposing  all  the  terms  to  one  member.  It  will  generally  be 
more  convenient  to  place  the  terms  which  do  not  contain  x  in 
the  opposite  member  from  those  which  are  multiplied  by  it. 

Example.    Let  the  equation  be 

mx  -\-  a  =1  nx  -\-  l,  (I) 

We  may  begin  by  transposing  a  to  the  second  member  and 
nx  to  the  first,  giving  at  once, 

mx  —  nx  ^=  h  —  a, 

or  (m  —  n)x  =z  b  -—  a, 

without  reducing  to  the  normal  form.  The  final  result  is  the 
same,  whatever  course  we  adopt,  and  the  division  of  both 
members  hjm  —  n  gives 

h  —  a 

7n  —  71 

139.  The  rule  which  may  be  followed  in  solving  equations 
of  the  first  degree  with  one  unknown  quantity  is  this : 

I.  Clear  the  equation  of  fractions, 

II.  Transpose  the  terms  which  are  multiplied  hy  the 
unhnown  quantity  to  one  member ;  those  which  do  not 
contain  it  to  the  other, 

III.  Divide  hy  the  total  coefficient  of  the  unhnown 
quantity. 


96  EQUATIONS   OF   THE  FIRST  DEGBEE. 

Note.  Rules  in  Algebra  are  given  only  to  enable  the  beginner  to  go 
to  work  in  a  way  which  will  always  be  sure,  though  it  may  not  always 
be  the  shortest.  In  solving  equations,  he  should  emancipate  liimself 
from  the  rules  as  soon  as  possible,  and  be  prepared  to  solve  each  equa- 
tion presented  by  such  process  as  appears  most  concise  and  elegant.  No 
operation  upon  the  two  members  in  accordance  with  the  axioms  (§  118) 
can  lead  to  incorrect  results  (provided  that  no  quantity  which  becomes 
zero  is  used  as  a  multiplier  or  divisor),  and  the  student  is  therefore  free 
to  operate  at  his  own  pleasure  on  every  equation  presented. 

EXAMPLES. 

^.  ax        . 

1.  Given  ^  =  1. 

hy 

It  is  required  to  find  the  value  of  each  of  the  quantities  a, 
h,  X,  and  y,  in  terms  of  the  others. 
Clearing  of  fractions,  we  have 

ax  =  hy. 
To  find  a,  we  divide  by  x,  which  gives 

III 
a  —  —' 

x 

To  find  ^,  we  divide  by  y,  which  gives 
ax  _ 

~y  ~ 

To  find  X,  we  divide  by  a,  which  gives 

by 
a 

To  find  y,  we  divide  by  d,  which  gives 
ax 

Thus,  when  any  three  of  the  four  quantities  a,  h,  x,  and  y, 
are  given,  the  fourth  can  be  found. 

2.  Let  us  take  the  equation, 

x  —  'i   —  ^^  +  ^ 

^X-\-  10    "~    A:X  —  % 

Clearing  of  fractions,  we  have 

4t2  —  30.^  -j-  14  =  4:«;2  _|_  32^^  4.  60. 


ONE    UNKNOWN  QUANTITY 


97 


» 


Transposing  and  reducing, 

—  %^x  —  46. 
Dividing  both  members  by  •—  62, 
_       46  _  _  46 

""■  ~  -  62  ~  ~  62 


23 

31' 


This  result  should  now  be  proved  by  computing  the  value  of  both 

23 
members  of  the  original  equation  when  —  —  is  substituted  for  a*. 

ol 

X       X  __ax       1 
^*     m       n~    h       m 

Proceeding  in  the  regular  way,  we  clear  of  fractions  by 
multiplying  by  mnh     This  gives 

nbx  +  7nl)X  =  amnx  —  nh. 

Transposing  and  reducing, 

{nd  -f-  md  —  amn)  x  =z  —  nh. 

Dividing  by  the  coefficient  of  x, 

__  nh  _  nb 

nb  -^  mb  —  amn       aynn  —  mb  —  nb 

These  two  values  are  equivalent  forms  (§  100). 

But  we  can  obtain  a  solution  without  clearing  of  fractions. 
ax 


Transposing  -y,  we  have 


X  X 

m       n 


ax 
T 


which  may  be  expressed  in  the  form 

/I        1       a\ 
\m       n       b! 


1^ 

m 


Dividing  by  the  coefficient  of  x, 

m 


X  ^=z   — 


1       ^_? 
m       71       b 


This  expression  can  be  redi  ced  to  the  other  by  §  110. 

7 


98  EQUATIONS   OF   THE  FIRST  DEGREE. 

EXERCISES. 

Find  the  values  of  x,  y,  or  u  in  the  following  equations : 
b  —  ^x       Sx  —  % 


X. 

3-3 

3- 

•^  4.  ^'  ^  ^  -  22 
i  +  2  +  3  -  ^^• 

S- 

y ^y    y -1 

a^  b       c~ 

7- 

U         U         tl- 

3-4  +  5=='^- 

9- 

u       u        1       1 

2. 

—  X  :=  a. 

4. 

a; +  23  _ 
x-^1   -^• 

6. 

36  _  45 
u  —  b~  u 

u  —  2Q.       8.     a—  dx  =:  b  -}-  ax, 
10.     3a;  H —  =  X. 


a      c 

c  —  x~~  a  —  X 


12. 


x  —  1      x  —  2       X  —  6      X ■ 


x^2    x  —  d     x  —  Q    x  —  n 

13-     —y  =  a  —  h. 

1  11  1 


14. 


x  —  2       X  —  4:        x  —  Q        X  —  1 


^5-  l(^-i)-|("--I)+i(^-i)  =  <>- 


tl  u  a 

16.     -  + 


a      h  —  a       h  -\-  a 

X        1 
17.     ax  -{-  b  =  -  +  y 

a       b       ■ 

u  —  a      u  —  b      u  —  c  _  u  —  {a  ■\-  b  -\-  c) 

b  c  a      ~~  abc 

711  (x  -\-  a)       n  (x  -\-  b) 

X  -\-b  X  -{-  a 

20.  (.^— ^)3  4-(.T_Z>)3  4.  {x—cf  =z  3  (:^;— «)  (x—b)  {x—c). 

Find  the  values  of  each  of  the  four  quantities,  a,  b,  c,  and 
d,  in  terms  of  the  other  three,  from  the  equations 

21.  7 h  T -7  =0.  22.       —  +  1  =r:  0. 

b  —  c       b  —  d  cd 


ONE    UNKNOWN  QUANTITY.  99 

Problems  leading  to  Simple  Equations. 

130.  The  first  difficulty  which  the  beginner  meets  with  in 
the  solution  of  an  algebraic  problem  is  to  state  it  in  the  form 
of  an  equation.  This  is  a  process  in  which  the  student  must 
depend  upon  his  own  powers.  The  following  is  the  general 
plan  of  proceeding : 

1.  Study  the  problem,  to  ascertain  what  quantities  in  it 
are  unknown.  There  may  be  several  such  quantities,  but  the 
problems  of  the  present  chapter  are  such  that  all  these  quan-^ 
titles  can  be  expressed  in  terms  of  some  one  of  them.  Select 
that  one  by  which  this  can  be  most  easily  done  as  the  unknown 
quantity. 

2.  Represent  this  unknown  quantity  by  any  algebraic  sym- 
bol whatever. 

It  is  common  to  select  one  of  the  last  letters  of  the  alpha- 
bet for  the  symbol,  but  the  student  should  accustom  himself 
to  work  equally  well  with  any  symbol. 

3.  Perform  on  and  with  these  symbols  the  operations  re- 
quired by  the  problem.  These  operations  are  the  same  that 
would  be  necessary  to  verify  the  adopted  value  of  the  unknown 
quantity. 

4.  Express  the  conditions  stated  or  implied  in  the  problem 
by  means  of  an  equation. 

5.  The  solution  of  this  equation  by  the  methods  already 
explained  will  give  the  value  of  the  unknown  quantity.  It  is 
always  best  to  verify  the  value  found  for  the  unknown  quan- 
tity by  operating  upon  it  as  described  in  the  equation. 

EXAMPLES. 

I.  A  sum  of  440  dollars  is  to  be  divided  among  three  people 
so  that  the  share  of  the  second  shall  be  30  dollars  more  than 
that  of  the  first,  and  the  share  of  the  third  80  dollars  less  than 
those  of  the  first  and  second  together.  What  is  the  share  of 
each? 

Solution.  1.  Here  there  are  really  three  unknown  quantities,  but 
it  is  only  necessary  to  represent  the  share  of  the  first  by  an  unknown 
symbol. 


100  EQUATIONS   OF   THE  FIB8T  DEGREE, 

2    Therefore  let  us  put 

X  =  share  of  the  iSrst. 

3.  Then,  by  the  terms  of  the  statement,  the  share  of  the  second  will  he 

X  +  30. 
To  find  the  share  of  the  third  we  add  these  two  together,  which  makes 

2x  +  30. 
Subtracting  80,  we  have 

2x  —  50 
as  the  share  of  the  third. 

We  now  add  the  three  shares  together,  thus, 

Share  of  first,  x 

"       "  second,       x-^-Z^ 

"       "  third,       2^--_50 
Shares  of  all,  ^x  —  20 

4.  By  the  conditions  of  the  problem,  these  three  shares  must  together 
make  up  440  dollars.  Expressing  this  in  the  form  of  an  equation,  we 
have 

^x  —  20  z=  440. 

5.  Solving,  we  find 

x  :=!  115  —  share  of  first. 
Whence,  115  -f  30  =r  145  =  share  of  second. 

115  +  145  —  80  =  J^O  =:  share  of  third. 
Sum  =  440.     Proof. 

Ex.  2.  Divide  the  number  90  into  four  parts,  such  that 
the  first  increased  by  2,  the  second  diminished  by  2,  the  third 
multiplied  by  2,  and  the  fourth  divided  by  2,  shall  all  be  equal 
to  the  same  quantity. 

Here  there  are  really  five  unknown  quantities,  namely,  the  four  parts 
and  the  quantity  to  which  they  are  all  to  be  equal  when  the  operation  of 
adding  to,  subtracting,  etc.,  is  performed  upon  them.  It  will  be  most 
convenient  to  take  this  last  as  the  unknown  quantity.  Let  us  therefore 
put  it  equal  to  u.     Then, 

Since  the  first  part  increased  by  2  must  be  eqaal  to  ?/,  its  value  will 
be  u  —  3. 

Since  the  second  part  diminished  by  2  must  be  equal  to  u,  its  value 
will  be  \i  +  2. 

Since  the  third  part  multiplied  by  2  must  be  \i,  its  value  will  be  ^  • 

Since  the  fourth  part  divided  by  2  must  make  u,  its  value  will  be  2?i. 


ONE    UNKNOWN  QUANTITY,  101 


Adding  these  four  parts  up,  their  sum  is  tbund  i;o  oe  --  • 

By  the  conditions  of  the  problem,  this  sum  must  make  up  the  num- 
Iber  90.     Therefore  we  have 

? = - 

Solving  this  equation,  we  find 

U  =  20. 
Therefore 

1st  part  =  u  —  2  =  IS. 

2d      ''     =:u  +  2  =  22. 

^  dd     ''     =u-T-2  =  10. 

4th    "     =2u        =  40. 

The  sum  of  the  four  equals  90  as  required,  and  the  first  part  increased 
by  2,  the  second  diminished  by  2,  etc.,  all  make  the  number  20,  as  re- 
quired. 


r 


PROBLEMS    FOR    EXERCISE. 

1.  What  number  is  that  from  which  we  obtain  the  same 
result  whether  we  multiply  it  by  4  or  subtract  it  from  100? 

2.  What  number  is  that  which  gives  the  same  result  when 
we  diyide  it  by  8  as  when  we  subtract  it  from  81  ? 

3.  Divide  284  dollars  among  two  people  so  that  the  share 
of  the  first  shall  be  three  times  that  of  the  second  and  $1G 
more. 

4.  Find  a  number  such  that  ^  of  it  shall  exceed  \  of  it 
by  12. 

5.  A  shepherd  describes  the  number  of  his  sheep  by  saying 
that  if  he  had  10  sheep  more,  and  sold  them  for  5  dollars  each, 
he  would  have  6  times  as  many  dollars  as  he  now  has  sheep. 
How  many  sheep  has  he  ? 

6.  An  iapplewoman  bought  a  number  of  apples,  of  which 
60  proved  to  be  rotten.  She  sold  the  remainder  at  the  rate  of 
2  for  3  cents,  and  found  that  they  averaged  her  one  cent  each 
for  the  whole.     How  many  had  she  at  first  ? 

7.  If  you  divide  my  age  10  years  hence  by  my  age  20  years 
ago,  you  will  get  the  same  quotient  as  if  you  should  divide  my 
present  age  by  my  age  26  years  ago.     What  is  my  present  age  ? 

8.  Divide  $500  among  A,  B,  and  0,  so  that  B  shall  have 
$20  less  than  A,  and  C  $20  more  than  A  and  B  together. 


102  EQUATlOm   OF   THE  FIRST  DEGREE. 

9.  A- father -left  $10000  to  be  diyidecl  among  his  five  chil- 
dren, directing  thai  each -should  receive  $500  more  than  the 
next  younger  one.     What  was  the  share  of  each  ? 

10.  A  man  is  6  years  older  than  his  wife.  After  they  have 
been  married  12  years,  8  times  her  age  would  make  7  times 
his  age.     What  was  their  age  when  married  ? 

11.  Of  three  brothers,  the  youngest  is  8  years  younger  than 
the  second,  and  the  eldest  is  as  old  as  the  other  two  together. 
In  10  years  the  sum  of  their  ages  will  be  120.  What  are  their 
present  ages  ? 

12.  The  head  of  a  fish  is  9  inches  long,  the  tail  is  as  long 
as  the  head  and  half  the  body,  and  the  body  is  as  long  as  the 
head  and  tail  together.     What  is  the  whole  length  of  the  fish  ? 

13.  In  dividing  a  year's  profits  between  three  partners,  A, 
B,  and  0,  A  got  one-fourth  and  $150  more,  B  got  one-third 
and  $300  more,  and  C  got  one-fifth  and  $60  more.  What  was 
the  sum  divided  ? 

14.  A  traveller  inquiring  the  distance  to  a  city,  was  told 
that  after  he  had  gone  one-third  the  distance  and  one-third 
the  remaining  distance,  he  would  still  have  36  miles  more  to 
go.     What  was  the  distance  of  the  city  ? 

15.  In  making  a  journey,  a  traveller  went  on  the  first  day 
one-fifth  of  the  distance  and  8  miles  more  ;  on  the  Second  day 
he  went  one-fifth  the  distance  that  remained  and  15  miles 
more ;  on  the  third  day  he  went  one-third  the  distance  that 
remained  and  12  miles  more  ;  on  the  fourth  he  went  35  miles 
and  finished  his  journey.  What  was  the  whole  distance 
travelled  ? 

16.  When  two  partners  divided  their  profits,  A  had  twice 
as  much  as  B.  If  he  paid  B  $300,  he  would  only  have  half  as 
much  again  as  B  had.     What  was  the  share  of  each  ? 

17.  At  noon  a  ship  of  war  sees  an  enemy's  merchant  vessel 
15  miles  away  sailing  at  the  rate  of  6  miles  an  hour.  How  fast 
must  the  ship  of  war  sail  in  order  to  get  within  a  mile  of  the 
vessel  by  6  o'clock  ? 

18.  A  train  moves  away  from  a  station  at  the  rate  of  h 
miles  an  hour.  Half  an  hour  afterward  another  train  follows 
it,  running  m  miles  an  hour.  How  long  will  it  take  the  latter 
to  overtake  it  ? 

19.  What  two  numbers  are  they  of  which  the  difference  is 
9,  and  the  difference  of  their  squares  351  ? 

20.  A  man  bought  25  horses  for  $2500,  giving  $80  a  piece 


ONE    UNKNOWN  QUANTITY  103 

for  poor  horses  and  $130  each  for  good  ones.     How  many  of 
each  kind  did  he  buy  ? 

21.  A  man  is  5  years  older  than  his  wife.  In  15  years  the 
snms  of  their  ages  will  be  three  times  the  present  age  of  the 
wife.     What  is  the  age  of  each  ? 

22.  How  far  can  a  person  who  has  8  hours  to  spare  ride  in 
a  coach  at  the  rate  of  6  miles  an  hour,  so  that  he  can  return  at 
the  rate  of  4  miles  ah  hour  and  arrive  home  in  time  ? 

23.  A  working  alone  can  do  a  piece  of  work  in  15  days, 
and  B  alone  can  perform  it  in  12  days.  In  what  time  can  they 
perform  it  if  both  work  together  ? 

Method  of  Solution.  In  one  day  A  can  do  ^^  of  the  whole  work 
and  B  can  do  yi-.     Hence,  both  together  can  do  (xV  +  tV)  of  it. 

If  both  together  can  do  it  in  x  days,  then  they  can  do  -  of  it  in  1  day. 

X 

XT  111 

Hence,  -  =  —  +  — 

X       13      15 

is  the  equation  to  be  solved. 

24.  A  cistern  can  be  filled  in  12  minutes  by  two  pipes  which 
run  into  it.  One  of  them  alone  will  fill  it  in  20  minutes.  In 
what  time  wotild  the  other  one  alone  fill  it  ? 

25.  A  cistern  can  be  emptied  by  three  pipes.  The  second 
pipe  runs  twice  as  much  as  the  first,  and  the  third  as  much  as 
the  first  and  second  together.  All  three  together  can  empty 
the  cistern  in  one  hour.  In  what  time  would  each  one  sepa- 
rately empty  it  ? 

26.  A  marketwoman  bought  apples  at  the  rate  of  5  for  two 
cents,  and  sold  half  of  them  at  2  for  a  cent  and  the  other  half 
at  3  for  a  cent.  Her  profits  were  50  cents.  How  many  did 
she  buy  ? 

27.  A  grocer  having  50  pounds  of  tea  worth  90  cents  a 
pound,  mixed  with  it  so  much  tea  at  60  cents  a  pound  that 
the  combined  mixture  was  worth  70  cents.  How  much  did 
he  add  ? 

28.  A  laborer  was  hired  for  40  days,  on  the  condition  that 
every  day  he  worked  he  should  receive  $1.50,  but  should  for- 
feit 50  cents  for  every  day  he  was  idle.  At  the  end  of  the 
time  $52  were  due  him.     How  many  days  was  he  idle  ? 

29.  A  father  left  an  estate  to  his  three  children,  on  the 
condition  that  the  eldest  should  be  paid  $1200  and  the  second 
$800  for  services  they  had  rendered.  The  remainder  was  to  be 
equally  divided  among  all  three.     Under  this  arrangement. 


104  EQUATIONS   OF   THE  FIRST  DEGREE. 

the  youngest  got  one-fourth  of  the  estate.     What  was  the 
amount  divided  ? 

30.  A  person  having  a  sum  of  money  to  divide  among 
three  people  gave  the  first  one-third  and  $20  more,  the  second 
one-third  of  what  was  left  and  $20  more,  and  the  third  one- 
third  of  what  was  then  left  and  $20  more,  which  exhausted  the 
amount.     How  much  had  they  to  divide  ? 

31.  One  shepherd  spent  $720  in  sheep,  and  another  got  the 
same  number  of  sheep  for  $480,  paying  $2  a  piece  less.  What 
price  did  each  pay? 

32.  A  crew  which  can  pull  at  the  rate  of  9  miles  an  hour, 
finds  that  it  takes  twice  as  long  to  go  up  the  river  as  to  go 
down.     At  what  rate  does  the  river  flow  ? 

33.  A  person  who  possesses  $12000  employs  a  portion  of 
the  money  in  building  a  house.  Of  the  money  which  remains, 
he  invests  one-third  at  four  jier  cent,  and  the  other  two-thirds 
at  five  per  cent.,  and  obtains  from  these  two  investments  an 
annual  income  of  $392.     What  was  the  cost  of  the  house  ? 

34.  An  income  tax  is  levied  on  the  condition  that  the  first 
$600  of  every  income  shall  be  untaxed,  the  next  $3000  shall 
be  taxed  at  two  per  cent.,  and  all  incomes  in  excess  of  $3600 
shall  be  taxed  three  per  cent,  on  the  excess.  A  person  finds 
that  by  a  uniform  tax  of  two  per  cent,  on  all  incomes  he  would 
save  $200.     What  was  his  income  ? 

35.  At  what  time  between  3  and  4  o'clock  is  the  minute- 
hand  5  minutes  ahead  of  the  hour  hand  ? 

2,6.  One  vase,  holding  a  gallons,  is  full  of  water ;  a  second, 
holding  h  gallons,  is  full  ot  brandy.  Find  the  capacity  of  a 
dipper  such  that  whether  it  is  filled  from  the  first  vase  and  the 
water  removed  replaced  by  brandy,  or  filled  from  the  second 
vase  and  the  latter  then  filled  with  water,  the  strength  of  the 
mixture  will  be  the  same. 

37.  Divide  a  number  m  into  four  such  parts  that  the  first 
part  increased  by  a,  the  second  diminished  by  a,  the  third 
multiplied  by  a,  and  the  fourth  divided  by  a  shall  all  be  equal. 

38.  Divide  a  dollars  among  five  brothers,  so  that  each  shall 
have  n  dollars  more  than  the  next  younger. 

39.  A  courier  starts  out  from  his  station  riding  8  miles  an 
hour.  Four  hours  afterwards  he  is  followed  by  another  riding 
10  miles  an  hour.  How  long  will  it  require  for  the  second  to 
overtake  the  first,  and  what  will  be  the  distance  travelled  ? 

If  X  be  the  number  of  hours  required,  the  second  will  have  travelled 
X  hours  and  the  first  (aj  + 4)  hours  when  they  meet.  At  this  time  they 
must  have  travelled  equal  distances. 


ONE    UNKNOWN  QUANTITY.  105 

Problem  of  the  Couriers. 

Let  us  generalize  the  preceding  problem  thus  : 

131.  A  courier  starts  out  from  his  station  riding  c 
miles  an  hour ;  h  hours  later,  he  is  followed  by  another 
riding  a  miles  an  hour.  How  long  will  the  latter  he  in 
overtaking  the  first,  and  wha/t  ivill  he  the  distance  from 
the  point  of  departure. 

Let  us  put  t  for  the  time  required.  Then  the  first  courier 
will  have  travelled  {t-\-1i)  hours,  and  the  second  t  hours. 
Since  the  first  travelled  c  miles  an  hour,  his  whole  distance  at 
the  end  of  ^  +  ^  hours  will  be  (t  +  Ji)  c.  In  the  same  way,  the 
distance  travelled  by  the  other  will  be  at.  When  fhe  latter 
overtakes  the  former,  the  distances  will  be  equal ;  hence, 

at  =  c{t^  h).  (1) 

Solving  this  equation  with  respect  to  t,  we  find 

t  =  -^.  (2) 

a  —  c  .       \  / 

Multiplying  by  a  gives  us  the  whole  distance  travelled, 
which  is 

Distance  = 

a  —  c 

This  equation  solves  every  problem  of  this  kind  by  substi- 
tuting for  a,  c,  and  li  their  values  in  numbers  supposed  in  the 
problem.  For  example,  in  Problem  39,  we  supposed  a  =  10, 
c=zS,  h=^4:.     Substituting  these  values  in  equation  (2),  we 

find 

t  =  16, 

which  is  the  number  of  hours  required. 

To  illustrate  the  generality  of  an  algebraic  problem,  we 
shall  now  inquire  what  values  f  shall  have  when  we  make  dif- 
ferent suppositions  respecting  a,  c,  and  h. 

(1.)  Let  us  suppose  a  =  c,  or  a  —  c=zO^  that  is,  the  rates 
of  travelling  equal.    Then  equation  (2)  will  become 

.  _  ch 


106  EQUATIONS   OF   THE  FIRST  DEOBEE. 

an  expression  for  infinity  (§  112,  6),  showing  that  the  one  conriei 
would  never  overtake  the  other.     This  is  plain  enough.     But, 

(2. )  Let  us  supipose  that  the  second  courier  does  not  ride 

so  fast  as  the  first,  that  is,  a  less  than  c,  and  a  —  c  negative. 

ch 

Then  the  fraction  will  not  be  infinite,  but  will  be  nega- 

it  —  c 

tive,  because  it  has  a  positive  numerator  and  a  negative  denom- 
inator. It  is  plain  that  the  second  courier  would  never  overtake 
the  first  in  this  case  either,  because  the  latter  would  gain  on 
him  all  the  time  ;  yet  the  fraction  is  not  infinite. 

What  does  this  mean  ? 

It  means  that  the  problem  solved  by  Algebra  is  more  gen- 
eral, that  is,  involves  more  particular  problems  than  were 
implied  in  the  statement.  If  we  count  the  hours  after  the 
second  courier  set  out  as  positive,  then  a  negative  time  will 
mean  so  many  hours  before  he  set  out,  and  this  will  bring  out 
a  time  when,  according  to  our  idea  of  the  problem,  the  horses 
were  still  in  the  stable. 

The  explanation  of  the  difficulty  is  this.     Suppose  S  to  be 

the  point  from  which  the  couriers  started,  and  AB  the  road 

alonfif  which  they  travelled  from 

A  S  B 

S  toward  B.     Suppose  also  that        i——....-^— 

the  first   courier    started    out 

from  S  at  8  o'clock  and  the  second  at  12  o'clock.  By  the  rule 
of  positive  and  negative  quantities,  distances  towards  A  are 
negative.  Now,  because  algebraic  quantities  do  not  commence 
at  0,  but  extend  in  both  the  negative  and  positive  directions, 
the  algebraic  problem  does  not  suppose  the  couriers  to  have 
really  commenced  their  journey  at  S,  but  to  have  come  from 
the  direction  of  A,  so  that  the  first  one  passes  S,  without  stop- 
ping, at  8  o'clock,  and  the  second  at  12.  It  is  plain  that  if  the 
first  courier  is  travelling  the  faster,  he  must  have  passed  the 
other  before  reaching  S,  that  is,  the  time  and  distance  are 
both  negative,  just  as  the  problem  gives  them. 

The  general  principle  here  involved  may  be  expressed  thus : 
In  Algebra,  roads  and  journeys,  like  time,  have  no  begin- 
ning  and  no  end. 


ONE    UNKNOWN  QUANTITY.  /lO? 

(3.)  Let  us  suppose  that  the  couriers  start  out  at  the  same 
time  and  ride  with  the  same  speed.  Then  li  and  a  —  c  are 
both  zero,  and  the  expression  for  t  assumes  the  form, 

'=1- 

This  is  an  expression  which  may  have  one  vahie  as  well  as 
another,  and  is  therefore  indeterminate.  The  result  is  correct, 
because  the  couriers  are  always  together,  so  that  all  values  of 
t  are  equally  correct. 

The  equation  (1)  can  be  used  to  solve  the  problem  iu  other 
forms.  In  this  equation  are  four  quantities,  a^  c,  h,  and  t,  and 
when  any  three  of  these  are  given,  the  fourth  can  be  found. 
There  are  therefore  four  problems,  all  of  which  can  be  solved 
from  this  equation. 

First  Problem,  that  already  given,  in  which  the  time 
required  for  one  courier  to  overtake  the  other  is  the  unknown 
quantity. 

Secokd  Problem.  A  courier  sets  out  from  a  station, 
riding  e  miles  an  hour.  After  h  hours  another  follows 
him  from  the  same  station,  intending  to  overtake  him 
ill  t  hours.    How  fast  must  he  ride  ? 

The  problem  can  be  put  into  the  form  of  an  equation  in 
the  same  way  as  before,  and  we  shall  have  the  equation  (1), 
only  a  will  now  be  the  unknown  quantity.  If  we  use  the 
numbers  of  Prob.  39  instead  of  the  letters,  we  shall  have,  in- 
stead of  equation  (1),  the  following  : 

16a  =:  8  (16  +  4)  =  8-20  =  160, 

whence  a  =  10. 

If  we  use  letters,  we  find  from  (1), 

.(^  +  70 

and  the  problem  is  solved  in  either  case. 

Third  Problem.  TJxe  second  courier  ca,n  ride  just  a 
miles  an  hour,  and  the  first  courier  starts  out  h  hours 


108  EQUATIONS   OF   THE  FIRST  DEGREE, 

before  him.     How  fast  must  the  latter  ride  in  order  that 
the  other  maij  take  t  hours  to  overtake  hiin  ? 

Here  c,  the  rate  of  the  first  courier,  is  the  unknown  quan- 
tity, and  by  solving  equation  (1),  we  find 

at 

Fourth  Problem.  The  swiftest  of  two  couriers  can 
ride  a  miles  an  hour,  and  the  slower  c  miles  an  hour. 
How  long  a  start  must  the  latter  have  in  order  that  the 
other  m^ay  require  t  hours  to  overtake  him  ? 

Here,  in  equation  (1),  h  is  the  unknown  quantity.  By 
solving  the  equation  with  respect  to  /?,  we  find, 

,        at  —  ct 

11  =  —-, 

which  solves  the  problem. 

PROBLEMS    OF    CIRCULAR    MOTION. 

40.  Two  men  start  from  the  same  point  to  run  repeatedly 
round  a  circle  one  mile  in  circumference.  If  A  runs  7  miles 
an  hour  and  B  5,  it  is  required  to  know : 

1.  At  what  intervals  of  time  will  A  pass  B  ? 

2.  At  how  many  different  points  on  the  circle  will  they  be 
together  ? 

We  reason  thus  :  since  A  runs  2  miles  an  hour  faster  than  B,  he  c:ets 
away  from  him  at  the  rate  of  2  miles  an  hour.  When  he  overtakes  him, 
he  will  have  gained  upon  him  one  circumference,  that  is,  1  mile.  This 
will  require  30  minutes,  which  is  therefore  the  required  interval.  In 
this  interval  A  will  have  gone  round  8^  and  B  2^  times,  so  that  they  will 
be  together  at  the  point  opposite  that  where  they  were  together  30 
minutes  previous.  Hence,  they  are  together  at  two  opposite  points  of 
the  circle. 

41.  What  would  be  the  answer  to  the  preceding  ques- 
tion if  A  should  run  8  miles  an  hour,  and  B  5  ? 

42.  Two  race-horses  run  round  and  round  a  course,  the 
one  making  tlie  circuit  in  30,  the  other  in  35  seconds.  If 
they  start  out  together,  how  long  before  they  will  be 
together  again  ? 

Note.    In  x  seconds  one  will  make  j^  circuit  and  the  other  jr^. 

o\j  00 

43.  If  one  planet  revolves  round  the  sun  in  T  and  the 
other  in  T'  years,  what  will  be  the  interval  between  their 
coni  unctions? 


TWO    UNKNOWN  QUANTITIES,  109 


CHAPTER    III. 

EQUATIONS   OF   THE    FIRST    DEGREE    WITH    SEVERAL 
UNKNOWN    QUANTITIES. 


Case  I.  Equations  with  Two  Unknown  Qiian- 
titles. 

133.  Def.  An  equation  of  the  first  degree  with  two 
unknown  quantities  is  one  which  admits  of  being  re- 
duced to  the  form 

ax  -{-by  =  6*5 

in  which  x  and  y  are  the  unknown  quantities  and  a,  &, 
and  c  represent  any  numbers  or  algebraic  equations 
which  do  not  contain  either  of  the  unknown  quantities. 

Def.  A  set  of  several  equations  containing  the  same 
unknown  quantities  is  called  a  System  of  Simulta- 
neous Equations. 

Solution  of  a  Pair  of  Simultaneous  Equations 
containing^  Two  Unknown  Quantities. 

133.  To  solve  two  or  more  simultaneous  equations, 
it  is  necessary  to  combine  them  in  such  a  way  as  to 
form  an  equation  containing  only  one  unknown  quan- 
tity. 

134.  Def.  The  process  of  combining  equations  so 
that  one  or  more  of  the  unknown  quantities  shall  dis- 
appear is  called  Elimination. 

The  term  "elimination"  is  used  because  the  unknown 
quantities  which  disappear  are  eliminated. 

There  are  three  methods  of  ehminating  an  unknown  quan- 
tity from  two  simultaneous  equations. 


110  EQUATIONS   OF   THE   FIRST  DEGREE. 

Elimination  by  Comparison. 

135.  Rule.  Solve  each  of  the  equations  with  respect 
to  one  of  the  unhnown  quantities  and  put  the  two  values 
of  the  unhnown  quantity  thus  obtained  equal  to  each 
other. 

This  will  give  an  equation  with  only  one  unhnown 
quantity,  of  which  the  value  can  he  found  from  the 
equation. 

The  value  of  the  other  unhnown  quantity  is  then 
found  hy  substitution. 


Example.     Let  the  equations  be 

ax  -{-  hy  ^^  c,   \ 
ax  +  h'y  =  c\  \ 

From  the  first  equation  we  obtain, 


(1) 


.'.  =  ^-^.  (2) 


(3) 


a 
From  the  second  we  obtain, 

c'  —  h'y 
a 

Putting  these  two  values  equal,  we  have 
c  —  hy  _  c'  —  h'y 


Eeducing  and  solving  this  equation  as  in  Chapter  II,  we 

find, 

_  ac'  —  a'c 

which  is  the  required  value  of  y.  Substituting  this  value  of  y 
in  either  of  the  equations  (1),  (2),  or  (3),  and  solving,  we  shall 
find 

_  Vc  -  he' 

~~  ah'  —  a'h 
If  the  work  is  correct,  the  result  will  be  the  same  in  which- 
ever of  the  equations  we  make  the  substitution. 


TWO    UNKNOWN  QUANTITIES.  Ill 

Numerical  Example.     Let  the  equations  be 

^  +  ^  =  28,) 
3x  -2ij  ^  29.  j  ^^ 

From  the  first  equation  we  find 

and  from  the  second  x  = — ^, 

o 

from  which  we  have  28  —  y  = — -, 

^  =  11. 
Substituting  this  value  in  the  first  equation  in  x,  it  becomes 

a:  =  28  -  11  =  17. 
If  we  substitute  it  in  the  second,  it  becomes 

29  +  22       51 

X  =  -3—  -  y  -  n, 

the  same  value,  thus  proving  the  correctness  of  the  work. 

Elimination  by  Substitution. 

136.  Rule.  Fijtd  the  value  of  one  of  the  unknoivn 
quantities  in  terms  of  the  other  from  either  equation, 
and  substitute  it  in  the  other  equation.  The  latter  will 
have  hut  one  unhnown  quantity. 

Example.    Taking  the  same  equations  as  before, 

ax  -^llj  r:^  c, 

a'x  ■\-  y  ij  =  c', 

the  first  "equation  gives        x  =  — — -* 

Substituting  this  value  instead  of  x  in  the  second  equation, 
it  becomes 

a'c  —  a'biJ      1,  , 

a  -^ 

Solving  this  equation  with  respect  to  y,  W^  get  the  same 
result  as  before. 


112  EQUATIONS   OF   THE  FIRST  DEOBEE. 

Numerical  Example.  To  solve  in  this  way  the  last  nu- 
merical example,  we  have  from  the  first  equation  (4), 

X  =  2%  —  y. 

Substituting  this  value  in  the  second  equation,  it  becomes 

84  —  3y  —  2y  =  29, 

from  which  we  obtain  as  before, 

84-^29 
y  =  — ^-  :=  11. 

This  method  may  be  applied  to  any  pair  of  equations  in 
four  ways  : 

1.  Find  X  from  the  first  equation  and  substitute  its  value 
in  the  second. 

2.  Find  x  from  the  second  equation  and  substitute  its 
value  in  the  first. 

3.  Find  y  from  the  first  equation  and  substitute  its  value 
in  the  second. 

4.  Find  y  from  the  second  equation  and  substitute  its 
value  in  the  first. 

Elimination  by  Addition  or  Subtraction. 

137.  EuLE.  Multiply  each  equation  hy  such  a  factor 
that  the  coefficients  of  one  of  the  unknown  quantities 
shall  become  numerically  equal  in  the  two  equations. 

Then,  by  adding  or  subtracting  the  equations,  we 
shall  have  an  equation  luith  but  one  unknown  quantity. 

Rem.  We  may  always  take  for  the  factor  of  each  equation 
the  coefficient  of  the  unknown  quantity  to  be  eliminated  in  the 
other  equation. 

Example.    Let  us  tako  once  more  the  general  equation 

ax  -}-  by  =  c, 
a'x  +  b'y  =z  c\ 
Multiplying  the  first  equation  by  «',  it  becomes 

aa!x  +  a!hy  =  a!c. 
Multiplying  tho  second  by  a,  it  becomes 
aa'x  +  aVy  =  ad. 


TWO    UNKNOWN   QUANTITIES.  113 

The  unknown  quantity  x  has  the  same  coefficient  in  the 

last  two  equations.     Subtracting  them  from  each  other,  we 

obtain 

{ah  —  ah') y  z=.  a'c  —  ac\ 

a'c  —  ac' 

y 


ah  —  ah' 


Rem.     We  shall  always  obtain  the  same  result,  whichever 
the  above  three  methods  we  use.     But  as 
last  method  is  the  most  simple  and  elegant. 


of  the  above  three  methods  we  use.     But  as  a  general  rule  the 


Problem  of  the  Sum  and  Difference. 

The  following  simple  problem  is  of  such  wide  application 
that  it  should  be  well  understood. 

138.  Problem.  The  sum  and  difference  of  two  num- 
bers hein^  given,  to  find  the  numbers. 

Let  the  numbers  be  x  and  y. 

Let  ,s*  be  their  sum  and  d  their  difference. 

Then,  by  the  conditions  of  the  problem, 

x  +  y  =  s, 
X  —  y  =i  d. 
Adding  the  two  equations,  we  have 
'^x  =1  s  +  d. 
Subtracting  the  second  from  the  first, 

^y  z=z  s  —  d. 
Dividing  these  equations  by  2, 

_  s  -\-  d  __  s       d 
^  -*  ~2"~  -2  +  2 
__  s  —  d  __  s       d 
y  ""  ~"2~  "^  2  ■"  2* 
We  therefore  conclude : 

The  greater  number  is  found  by  adding  half  the  dif- 
ference to  half  the  sum. 

The  lesser  nuynber  is  found  by  subtracting  half  the 
difference  front'  half  the  sum. 


114  EQUATIONS   OF   THE  FIMST  DEGREE. 

This  result  can  be  illustrated  geometrically.     Let  AB  and 
BC  be  two  lines  placed  end  to  end,  so  that  AC  is  their  sum. 
To  find  their  difference,  we 
cut   off  from  AB  a  length        ^  ^\  ^\  ^ 


AC  r=  BC  ;  then  C'B  is  the  i 

P 
difference  of  the  two  lines. 

If  P  is  half  way  between  C  and  B,  it  is  the  middle  point 

of  the  whole  line,  so  that 

AP  =  PC  =  iAC  =  i  sum  of  lines. 
CT  r=  PB  =  iC'B  =  i  difference  of  lines. 

If  to  the  half  sum  AP  we  add  the  half  difference  PB,  we 
have  AB,  the  greater  line. 

If  from  the  half  sum  AP  we  take  the  half  difference  C'P, 
we  have  left  AC,  the  lesser  line. 

EXERCISES. 

Solve  the  following  equations : 


I. 

Zx  —  2y 

=  33, 

2x  —  3y 

=  18. 

2. 

3x  —  6y 

-=13, 

2x  +  7y 

=  81. 

3- 

7x-{-  (jy 

=  a, 

Qx-\-6y 

=  b. 

4- 

2x  +  dy 

=  m, 

2x-'3y 

=  n. 

5- 

ax  +  by 

:=  p, 

ax  —  by 

=  5'- 

6. 

6"^  7 

=  26, 

X      y 
6  ""7 

=  2. 

7. 

=  18, 

^  .y 

8  "^2 

=  29. 

8. 

2^3 

==  a, 

X      y 
2   3 

=  b. 

9-     'i  (*•  +  «/)  +  3  {x  -  y)  =  103, 

Note.  Solve  this  equation  first  as  if  aj+^  and  x—y  were  single  sym- 
bols, of  which  the  values  are  to  be  found.  Then  find  x  and  y  hy  %  138 
preceding. 

lo.     x-^y  -\-  {x-^y)  =  U,    x  +  y  ^  ^c  -  y)  =  10. 
^^'     x'^  y^  12'    x      ^  ~  12* 


TWO    UNKNOWN  QUANTITIES.  115 

Note.    Equations  in  this  form  can  be  best  solved  as  if  -  and  -  were 
the  unknown  quantities.     See  next  exercise.  ^ 

^  '     X      y       1^^    X      y 

Solution.    If  we  multiply  the  first  equation  by  4,  and  the  second  by 
8,  we  have 

12_  8    _  44  _  22 

X         y   ~  10  ""   5  ' 

^1  +  ^A  =  ,  =  % 

X        y  5 

Subtracting  the  first  from^he  second,  we  have 
23  __  23 
y  -  b' 

whence, 

y  =  b. 

Again,  to  eliminate  - ,  we  multiply  the  first  equation  by  5  a»d  the 
second  by  3  and  add.    Thus, 

15  __  1 0  _  11 
X        y  ~  2' 

8  ^  10        „       12 


whence, 


23  _  23^ 

X  ~   2' 

X  =  2. 


2       3_^      ?_3___L 
^^'     x'^  y~  12'    X      y~       12 

1       2  _   b_      2_^_^ 
^^'     x'^  y''  12'    X      y~  24' 
5_3__1      3_1__1^ 
X       y~        Q'     X       y  ~  ^O' 

5      _     3      _  _  1      _3 1_  _   1^ 

x  +  1      y  —  l~~       6'    x-i-l      y  —  l'^SO' 

2  3      _   7  2 3      _         1 

^7-     x  +  2'^  y  —  d~'  12'    x-i'2~y--d'~'       12 


15. 
16. 


116  EQUATIONS  OF  THE  FIRST  DEGBEE. 

^      a      I  a      h        ^ 

X      y  ^      y 

Case  II.  Equatio7is  of  the  First  JDegree  with 
Three  or  More  Unknown  Quantities. 

139.  When  the  values  of  several  unknown  quantities  are 
to  be  found,  it  is  necessary  to  have  as  many  equations  as  un- 
known quantities. 

If  there  are  more  unknown  quantities  than  equations,  it 
will  be  impossible  to  determine  the  values  of  all  of  them  from 
the  equations.  All  that  can  be  done  is  to  determine  the  value 
of  some  in  terms  of  the  others. 

If  the  number  of  equations  exceeds  that  of  unknown  quan- 
tities, the  excess  of  equations  will  be  superfluous.  If  there 
are  n  unknown  quantities,  their  values  can  be  found  from  any 
n  of  the  equations.  If  any  selection  of  n  equations  we  choose 
to  make  gives  the  same  values  of  the  unknown  quantities,  the 
equations,  though  superfluous,  will  be  consistent.  If  different 
values  are  obtained,  it  will  be  impossible  to  satisfy  them  all. 

Elimination. 

140.  When  the  number  of  unknown  quantities  exceeds 
two,  the  most  convenient  method  of  elimination  is  generally 
that  by  addition  or  subtraction.  The  unknown  quantities  are 
to  be  eliminated  one  at  a  time  by  the  following  method  : 

I.  Select  an  unknown  quantity  to  he  first  eliminated. 
It  is  best  to  begin  ivith  the  quantity  which  appears  in 
the  fewest  equations  or  has  the  simplest  coejflcients. 

II.  Select  one  of  the  equations  containing  this  un^ 
Icnown  quantity  as  an  eliminating  equation. 

III.  Eliminate  the  quantitxj  between  this  equation 
and  each  of  the  others  in  succession. 


THREE   OB   MORE    UNKNOWJS^  QUANTITIES.       117 

We  shall  then  have  a  second  system  of  equations  less  bj 
one  in  number  than  the  original  system  and  containing  a  num- 
ber of  unknown  quantities  one  less. 

IV.  Repeat  the  process  on  the  new  system  of  equations, 
and  continue  the  repetition  until  only  one  equation  with 
one  unhnown  quantity  is  left. 

Y.  Having  found  the  value  of  this  last  unhnown 
quantity,  the  values  of  the  others  can  he  found  by  suc- 
cessive substitution  in  one  equation  of  each  system. 

Example.     Solve  the  equations 

(1)  42:  —  3y  —   2;  +    ^  —    7  =  0, 

(2)  X—   y -^2z-\-2u  —  10  =zO 

(3)  2a;  +  2y  —    z-^^ti—    2  =  0,^  ^^^ 

(4)  a;  +  2y  +    ;$;  +    2t  —  19  =  0. 

We  shall  select  x  as  tlie  first  quantity  to  be  eliminated,  and  take  the 
last  equation  as  the  eliminating  one.  We  first  multiply  this  equation  by 
three  such  factors  that  the  coefficient  of  x  shall  become  equal  to  the  co- 
efficient of  X  in  each  of  the  other  equations.  These  factors  are  4,  1,  and  2. 
We  write  the  products  under  each  of  the  other  equations,  thus  : 

Eq.  (1),  ^x-^^y  —    z-\-    u—    1  =  0, 

(4)  X  4,  4:r  4-  8y  +  ^^  +  ^u  —  76  ^  0. 

Eq.  (2),  X—    y  -\-'^z-\-2ii  —  10  =  0, 

(4)  X  1,  :g  -f  2y  +    ^  +    ^^  —  19  rzz  0. 

Eq.  (3),  2x-^2y  —    z  —  2u~-    2  =  0, 

(4)  X  2,  2a;  +  4^/  +  2^  +  ^u  —  38  =  0. 

By  subtracting  the  one  of  each  pair  from  the  other,  we  obtain  the 
equations, 

lly  -f  5;^  +  3i^  —  69  =  0,  ^ 

2y  -{-dz-\-  4:11  —  36  =  0.  ) 

The  unknown  quantity  x  is  here  eliminated,  and  we  have  three  equa- 
tions with  only  three  unknown  quantities.  Now  eliminating  y  by  means 
of  the  last  equation,  in  the  same  way,  and  clearing  of  fractions,  we  find 
the  two  equations, 


232;  +  38?^  —  258  =  0, 
llz  +  14?^  —    90  =  0, 


: }  (^) 


118  EQUATIONS   OF   THE  FIRST  DEGREE, 

The  problem  is  now  reduced  to  two  equations  with  two  unknown 
quantities,  which  we  have  already  shown  how  to  solve.  We  find  by 
solving  them, 

u  —  %, 

We  next  find  the  value  of  y  by  substituting  these  values  of  z  and  u 
in  either  of  the  equations  (b).    The  first  of  them  thus  becomes: 

lly  _  10  +  24  —  69  =  0, 

from  which  we  find, 

2^  =  5. 

We  now  substitute  the  values  of  y,  2,  and  u  in  either  of  equations  {a). 
The  second  of  the  latter  becomes 

iT  —  5  —  4  +  16  —  10  =  0, 

and  the  fourth  becomes, 

o;  +  10  —  2  +  8  —  19  =  0, 

either  of  which  gives 

o:  =  3. 

We  can  now  prove  the  results  by  substituting  the  values  of  a*,  y,  z, 
and  u  in  all  four  of  equations  {a)^  and  seeing  whether  they  are  all  satisfied. 

EXERCISES. 

1.  One  of  the  best  exercises  for  the  student  will  be  that  of 
resolving  the  previous  equations  {a)  by  taking  the  last  equa- 
tion as  the  eliminating  one,  and  performing  the  elimination 
in  different  orders;  that  is,  begin  by  eliminating  u,  then 
repeat  the  whole  process  beginning  with  z,  etc.  The  final 
results  will  always  be  the  same. 

2.  Find  the  values  of  x^,  x^^  x^,  and  x^,  from  the  equa- 
tions, 

i^l   +  ^2   +  ^3   +  ^4    =    ^^> 
X^    "T  ^2        '  *^3  *^4    "~~      ^ 

iV  -j     ^"^   U^  n     ~|~    %fj  q    "~~   »^  A      ———  V, 

This  example  requires  no  multiplication,  but  only  addition  and  sub 
traction  of  the  different  equations. 

3a  2x+6y  +  3z  =  13, 

2a;  +  2^  —    z  =  12, 
6x  +  5y--2z  =  29. 


PROBLEMS.  119 


dz  +  2u- 

•hy 

=  18, 

3x  -\-y  — 

Au 

-    9, 

X 

-\-lz- 

.6y 

=z  33, 

bz- 

-2x- 

-8y  + 

2u 

=  15. 

X  -{-  y  +  z 
y  +  z  +  u 

6. 

1  1 
X  y 
1       1 

=  m 

z  -}-  U  +  X 

1^  +  ^  +  y 

=z  C, 

y~  z 
1       1 

-z^x 

PROBLEMS    FOR    SOLUTION. 

1.  A  man  had  a  saddle  worth  $75  and  two  horses.  If  the 
saddle  be  put  on  horse  A  he  will  be  double  the  value  of  B,  but 
if  it  be  put  on  B  his  value  will  be  equal  to  that  of  A.  What 
is  the  value  of  each  horse  ? 

2.  What  number  of  two  digits  is  equal  to  7  times  the  sum 
of  its  digits,  and  to  9  times  the  difference  of  its  digits  increased 
by  4? 

Let  X  be  the  first  digit,  or  the  number  of  tens,  and  y  the  units.  Then 
the  number  itself  will  be  lOar+y.     Seven  times  the  sum  of  the  digits  are 

7aJ4-7^,  and  9  times  the  difference  is  9(aj— ^  +  4). 

3.  A  number  of  two  digits  is  equal  to  6  times  the  sum  of 
its  digits,  and  if  9  be  subtracted  from  the  number  the  digits 
are  reversed.    What  is  the  number? 

4.  Find  a  number  of  two  digits  such  that  it  shall  be  equal 
to  6  times  the  sum,  of  its  digits  increased  by  1,  while  if  18  be 
subtracted  from  the  number  the  digits  will  be  reversed. 

5.  Find  a  number  which  is  greater  by  2  than  5  times  the 
sum  of  its  digits,  and  if  9  be  added  to  it  the  digits  will  be 
reversed. 

6.  What  number  is  that  which  is  equal  to  9  times  the  sum 
of  its  digits  and  is  4  greater  than  11  times  their  diifereYice  ? 

7.  What  fraction  is  that  which  becomes  equal  to  f  when 
the  numerator  is  increased  by  2,  and  equal  to  f  when  the  de- 
nominator is  increased  by  4. 

8.  Two  drovers  A  and  B  went  to  market  with  cattle.  A 
sold  50  and  then  had  left  half-  as  many  as  B,  who  had  sold 
none.  Then  B  sold  54  and  had  remaining  half  as  many  as  A. 
How  many  did  each  have? 


120  EQUATIONS  OF  TEE  FIRST  DEGREE. 

9.  A  boy  bought  42  apples  for  a  dollar,  giving  3  cents  each 
for  the  good  ones  and  2  cents  each  for  the  poor  ones.  How 
many  of  each  kind  did  he  buy  ? 

10.  Find  a  fraction  which  becomes  equal  to  ^  when  its 
denominator  is  increased  by  13,  and  to  f  when  4  is  subtracted 
from  its  numerator. 

11.  Find  a  fraction  which  will  become  equal  to  f  by  adding 
2  to  its  numerator,  or  by  adding  to  its  denominator  3,  will  be- 
come ^. 

12.  A  huckster  bought  a  certain  number  of  chickens  at 
32  cents  each  and  of  turkeys  at  75  cents  each,  paying  $14  for 
the  whole.  He  sold  the  chickens  at  48  cents  each,  and  the 
turkeys  at  $1  each,  realizing  $20  for  the  whole.  How  many 
chickens  and  how  many  turkeys  had  he  ? 

13.  An  apple  woman  bought  a  lot  of  apples  at  1  cent  each, 
and  a  lot  of  pears  at  2  cents  each,  paying  $1.70  for  the  whole. 

11  of  the  apples  and  7  of  the  pears  were  bad,  but  she  sold  the 
good  apples  at  2  cents  each  and  the  good  pears  at  3  cents  each, 
realizing  $2.60.     How  many  of  each  fruit  did  she  buy? 

14.  When  Mr.  Smith  was  married  he  was  \  older  than  his 
wife  ;  twelve  years  afterward  he  was  |  older.  What  were  their 
ages  when  married.? 

15.  A  and  B  together  can  do  a  piece  of  work  in  6  days,  but 
A  working  alone  can  do  it  9  days  sooner  than  B  working 
alone.     In  what  time  could  each  of  them  do  it  singly  ? 

16.  A  husband  being  asked  the  age  of  himself  and  wife, 
replied:  ^^If  you  divide  my  age  6  years  hence  by  her  age 
6  years  ago,  the  quotient  will  be  2.     But  if  you  divide  her  age 

12  years  hence  by  mine  21  years  ago,  the  quotient  will  be  5. 

17.  The  sum  of  two  ages  is  9  times  their  difference,  but 
seven  years  ago  it  was  only  seven  times  their  difference.  What 
are  the  ages  now  ? 

18.  Two  trains  set  out  at  the  same  moment,  the  one  to  go 
from  Boston  to  Springfield,  the  other  from  Springfield  to  Bos- 
ton. The  distance  between  the  two  cities  is  98  miles.  They 
meet  each  other  at  the  end  of  1  hr.  24  min.,  and  the  train  from 
Boston  travels  as  far  in  4  lirs.  as  the  other  in  3.  What  was  the 
speed  of  each  train  ? 

19.  A  grocer  bought  50  lbs.  of  tea  and  100  lbs.  of  coffee  for 
$60.  He  sold  the  tea  at  an  advance  of  J  on  his  price,  and  the 
coffee  at  an  advance  of  \,  realizing  $77  from  both.  At  what 
price  per  pound  did  he  buy  and  sell  each  article  ? 

Note.  If  x  and  y  are  the  prices  at  which  he  bought,  then  \x  and  |y 
are  the  prices  at  which  he  sold. 


INCONSISTENT  EQUATIONS.  121 

20.  For  p  dollars  I  can  purchase  either  a  pounds  of  tea  and 
h  pounds  of  coffee,  or  m  pounds  of  tea  and  n  pounds  of  coffee. 
What  is  the  price  per  pound  of  each  ? 

21.  A  goldsmith  had  two  ingots.  The  first  is  composed  of 
equal  parts  of  gold  and  silver,  while  the  second  contains  5  parts 
of  gold  to  1  of  silver.  He  wants  to  take  from  them  a  watch- 
case  having  4  ounces  of  gold  and  1  ounce  of  silver.  How 
much  must  he  take  from  each  ingot  ? 

22.  A  banker  has  two  kinds  of  coin,  such  that  a  pieces  of 
the  first  kind  or  h  pieces  of  the  second  will  make  a  dollar.  If 
he  wants  to  select  c  pieces  which  shall  be  worth  a  dollar,  how 
many  of  each  kind  must  he  take  ? 

23.  A  has  a  sum  of  money  invested  at  a  certain  rate  of 
interest.  B  has  $1000  more  invested,  at  a  rate  1  per  cent, 
higher,  and  thus  gains  $80  more  interest  than  A.  C  has  in- 
vested 1500  more  than  B,  at  a  rate  still  higher  by  1  per  cent., 
and  thus  gains  $70  more  than  B.  What  is  the  amount  each 
person  has  invested  and  the  rate  of  interest  ? 

24.  A  grocer  had  three  casks  of  wine,  containing  in  all 
344  gallons.  He  sells  50  gallons  from  the  first  cask;  then 
pours  into  the  first  one-third  of  what  is  in  the  second,  and 
then  into  the  second  one-fifth  of  what  is  in  the  third,  after 
which  the  first  contains  10  gallons  more  than  the  second, 
and  the  second  10  more  than  the  third.  How  much  wine  did 
each  cask  contain  at  first  ? 

Equivalent  and  Inconsistent  Equations. 

141.  It  is  not  always  the  case  that  values  of  two  unknown 
quantities  can  be  found  from  two  equations.    If,  for  example, 

we  have  the  equations 

a;  4-  2^  =  3, 
2x  -f  4^  r=  6, 

we  see  that  the  second  can  be  derived  from  the  first  by  multi- 
plying both  members  by  2.     Hence  every  pair  of  values  of  x 
and  y  which  satisfy  the  one  will  satisfy  the  other  also,  so  that 
the  two  are  equivalent  to  a  single  one. 
If  the  equations  were 

a;  +  2y  =  5, 

2x-\-4.y  =  6, 
there  would  be  no  values  of  x  and  y  which  would  satisfy  both 
equation.s.  , 


122  EQUATIONS   OF   THE  FIRST  DEGREE, 

For,  if  we  multiply  the  first  by  2  and  subtract  the  second 
from  the  product,  we  shall  have, 

1st  eq.  X  2,  2:^  4-  %  =  10 

2d  eq.,  %x_^rj^y_j=_^ 

Remainder,  0=4, 

an  impossible  result,  which  shows  that  the  equations  are  incon- 
sistent. This  will  be  evident  from  the  equations  themselves, 
because  every  pair  of  values  of  x  and  y  which  gives 

2:c  +  %  =  6, 
must  also  give  ic  +  2^  ==  3, 

and  therefore  cannot  give   a;  +  2^  =  5. 

143.  Generalization  of  the  preceding  result.  If  we  take 
any  two  equations  of  the  first  degree  between  x  and  y  which 
we  may  represent  in  the  form 

ax-\-ly  —  c,   )  ,  . 

dx  +  Vy  =z  c',  \  ^^ 

and  eliminate  x  by  addition  or  subtraction,  as  in  §  137,  we  have 
for  the  equation  in  y, 

{ad  —  ab')  y  =:  a'c  —  ac', 
Now  it  may  happen  that  we  have, 

a'b  —  ab'  =  0  identically.  (2) 

In  this  case  y  will  disappear  as  well  as  x,  and  the  result 

will  be 

a'c  —  ac'  =  0. 

If  this  equation  is  identically  true,  the  two  equations  (1) 
will  be  equivalent ;  if  not  true,  they  will  be  inconsistent.  In 
neither  case  can  we  derive  any  value  of  y  or  x. 

If  we  divide  the  above  equation,  (2),  by  aa'  we  shall  have 

b  _  b^^ 

a  '~  a' 
Hence, 

Theorem.  If  the  quotient  of  the  coefficients  of  the 
unknown  quantities  is  the  same  in  the  tv^o  equations, 
they  will  be  either  equivalent  or  inconsistent. 


INEQUALITIES,  123 

This  theorem  can  be  expressed  in  the  following  form : 

If  the  terms  containing  the  unknown  quantity  in  the 
one  equation  can  he  multiplied  by  such  a  factor  that 
they  shall  both  become  equal  to  the  corresponding  terms 
of  the  other  equation,  the  two  equations  will  be  either 
equivalent  or  inconsistent. 

Proof,     If  there  be  such  a  factor  m  that  multiplying  the 
first  equation  (1)  by  it,  we  shall  have 
ma  =  a\ 
mb  =1  V. 
Eliminating  m,  we  find 

a!b  —  aV  =  0, 
the  criterion  of  inconsistency  or  equivalence. 

143.  When  two  equations  are  inconsistent,  there  are  no 
values  of  the  unknown  quantities  which  will  satisfy  both  equa- 
tions. 

When  they  are  equivalent,  it  is  the  same  as  if  we  had  a 
single  equation ;  that  is,  we  may  assign  any  value  we  please  to 
one  of  the  unknown  quantities,  and  find  a  corresponding  value 
of  the  other. 


CHAPTER     IV. 
OF     INEQUALITIES. 

144.  Def.  An  Inequality  is  a  statement,  in  the 
language  of  Algebra,  that  one  quantity  is  algebraically 
greater  or  less  than  another. 

Def.  The  quantities  declared  unequal  are  called 
Members  of  the  inequality. 

The  statement  that  A  is  greater  than  B,  or  that  A  —  B  is 
positive,  is  expressed  by 

A  >  B. 


124  INEQUALITIES. 

That  A  is  less  than  B,  or  that  A  —  B  i^  negative  is 
expressed  by 
^  ^  A<B, 

The  form  Ay  B  >  G 

indicates  that  the  quantity^  is  less  than  A  but  greater  than  Co 

The  form  A^  B 

indicates  that  A  may  be  either  equal  to  or  greater  than  B,  but 
cannot  be  less  than  B, 

Properties  of  Inequalities. 

145.  Theorem  I.  An  inequality  will  still  subsist 
after  the  same  quantity  has  been  added  to  or  subtracted 
from  each  member. 

Proof,  If  the  inequality  hQ  Ay  B,  A  —  B  must  be  posi- 
tive. If  we  add  the  same  quantity  H  to  A  and  B,  or  subtract 
it  from  them,  we  shall  have  A  ±^  H  —  {B  ±  H),  which  is 
equal  to  A  —  B,  and  therefore  positive.     Hence,  if  j 

A>  B, 

then  A±H>  B  ±E. 

Cor.  If  any  term  of  an  inequality  be  transposed 
and  its  sign  changed,  the  inequality  will  remain  true. 

Theorem  IL  An  inequality  will  still  subsist  after 
its  members  have  been  multiplied  or  divided  by  the 
same  positive  number. 

Proof.  If  ^  —  i?  is  positive,  then  (m  or  n  being  positive) 
m  {A  —  B)  or  mA  —  7nB  will  be  positive,  and  so  will 


A 

-^            A      B 

or 

n                  71       n 

Hence,  if 

A>  B, 

then 

mA  >  mB, 

and 

n        n 

INEQUALITIES.  125 

It  may  be  shown  in  the  same  way  that  if  m  or  n  is  negative, 

A       B 

mA  —  niB  or will  be  negative.     Hence, 

n       01  ^ 

Theorem  III,  If  botli  members  of  an  inequality  Tbe 
multiplied  or  divided  by  the  same  negative  number, 
the  direction  of  the  inequality  will  be  reversed. 

That  is,  if  A  >  B, 

then  —  mA  <  —  mB, 

and  < 

n  n 

Theorem  IV.  If  the  corresponding  members  of 
several  inequalities  be  added,  the  sura  of  the  greater 
members  will  exceed  the  sum  of  the  lesser  members. 

Theorem  F.  If  the  members  of  one  inequality  be 
subtracted  from  the  non-corresponding  members  of 
another,  the  inequality  will  still  subsist  in  the  direction 
of  the  latter. 

That  is,  if  Ay  B, 

then  A  —  y  y  B  —  X. 

The  proof  of  the  last  three  theorems  is  so  simple  that  it  may  be  sup- 
plied by  the  student. 

Theorem  VI  If  two  positive  members  of  an  in- 
equality be  raised  to  any  power,  the  inequality  will 
still  subsist  in  the  same  direction. 

Proof,     Let  the  inequality  be 

Ay  B.  {a) 

Because  A  is  positive,  we  shall  have,  by  multiplying  by  A 
(Th.  II), 

A^  >  AB,  (1) 

Also,  because  B  is  positive,  we  liave,  by  multiplying  (a) 
AB  y  &.  (2) 


126  INEQUALITIES. 

Therefore,  from  (1)  and  (2), 

A^  >  B\  (3) 

Multiplying  the  last  inequality  by  Ay 

A^  >  AB\  (4) 

Multiplying  (2)  by  ^, 

AB'^  >  B\  (5) 

Whence,  A^  >  B\ 

The  process  may  be  continued  to  auy  extent. 

Examples  of  the  Use  of  Inequalities. 

146.    Ex.  I.  If  a  and  h  be  two  positive  quantities,  such 
that 

«2  +  ^2  ^  1^ 

Tve  must  have  «  +  Z>  >  1. 

Proof,    If  a-\~h^l, 

we  should  have,  by  squaring  the  members  (Th.  VI), 
^2  _^  2a&  +  ^>2  =  1 . 

and  by  transposing  the  product  'Zah  (Th.  I,  Cor.), 
«2  +  Z>2  =  1  _  <^ad. 

Because  a  and  ^  are  positive,  "Hah  is  positive,  and 

1  _  2aZ>  <  1. 
Therefore  we  should  have 

a^  +  y^  <  1, 
and  could  not  have  a^  +  1)^  —1,  as  was  originally  supposed. 

Ex.  2.    If  flj,  Z>,  w,  and  ^^  are  positive  quantities,  such  that 
a  ^   m  ,  V 

b  >  ^'  ^«> 

then  the  value  of  the  fraction will  be  contained  between 

a  +  n 

ct  fH 

the  values  of  ^  and  —  ;   that  is, 
0  n 


(^) 


(3) 


WEQUALITIES.  127 

a       a  +  m       m 

b  ^  b-\-n   ^  n'  ^^ 

To  prove  the  first  inequality,  we  must  show  that 

a      a  +  m 
b      b  -\-  n 

is  positive.     Keducing  this  expression  by  §  106,  it  becomes 

an  —  hn 
b~{b~+n)' 

From  the  original  inequality  (a)  we  have,  by  multiplying 
by  the  positive  factor  bn, 

an  >  bm. 

That  is,  an  —  bm  is  positive ;  therefore  the  fraction  (3) 
with  this  positive  numerator  is  also  positive,  and  (2)  is  positive 
as  asserted. 

The  second  inequality  (1)  may  be  proved  in  the  same  way. 

EXERCISES. 

I.  Prove  that  if  a  and  b  be  any  quantities  different  from 
zero,  and  1  >  ic  >  —  1,  we  must  have 

«2  _  2abx  +  b^>  0. 

(a  +  bV 
— - — )  >  ab, 

3.  If  3a;  —  5  >  13,  then  x  >  6. 

4.  If  6a;  >  y  +  18,  then  a;  >  4. 

5.  If  ^-^>|-3,  thena;>5. 

6.  li  m  —  nx  y  p  —  qx,   then   x  > 

7.  If  "  <  1 '■,  and  m  and  y  of  like  sign :  x  <  y. 

'my 

8.  If  «2  _{_  J2  _^  ^2  _  1^  and  a,  b,  and  c  are  not  all  equal, 

then  ab  -\-  be  +  ca  <i  1. 

Suggestion.  The  squares  of  a  —  byb  —  c,  and  c  —  a  cannot  be 
negative. 


BOOK    IV. 
RATIO     AND     PROPORTION. 


CHAPTER    I. 
NATURE    OF    A     RATIO. 

14'7.  Def,  The  Ratio  of  a  quantity  A  to  another 
quantity  j8  is  a  number  expressing  the  value  of  A  when 
compared  with  B  as  the  standard  or  unit  of  measure. 

EXA.MPLES.     Comparing  ^    ^     |     |     |     |     |     |     |     |     | 

the  lengths  A,  B,  C,  D,  it  j— i— r 

will  be  seen  that  '     '     ' 

A  is  ^  times  D]  (^    1     \     ~\ 

B  18  i  of  D;  2>    I     I     I     I     I 
(7  is  I  of  D, 

We  express  this  relation  by  saying, 

9 
The  ratio  of  A  to  D  is  2^  or  -; 

''     ^  to  i>  is  ^;  )  (1) 

"  ''      C  to  D  is  ^' 

4 

148.  The  ratio  of  one  quantity  to  another  is  expressed  by 
writing  the  unit  of  measure  after  the  quantity  measured,  and 
inserting  a  colon  between  them. 

The  statements  (1)  will  then  be  expressed  thus  : 

A:D  =  2i=:l;      ^'•^  =  h       ^  '  ^  =  l 

Def.  The  two  quantities  compared  to  form  a  ratio 
are  called  its  Terms. 


RATIO.  129 

Def.  The  quantity  measured,  or  the  first  term  of 
the  ratio,  is  called  the  Antecedent. 

The  unit  of  measure,  or  the  second  term  of  the  ratio, 
is  called  the  Consequent. 

Eem.  When  the  antecedent  is  greater  than  the  consequent, 
the  ratio  is  greater  than  unity. 

When  the  antecedent  is  less  than  the  consequent,  the  ratio 
is  less  than  unity. 

149.  To  find  the  ratio  of  a  quantity  ^  to  a  standard  U, 
we  imagine  ourselves  as  measuring  off  the  quantity  A  with  Z7as 
a  carpenter  measures  a  board  with  his  foot-rule. 

There  are  then  three  cases  to  be  considered,  according  to 
the  way  the  measures  come  out. 

Case  I.  We  may  find  that,  at  the  end,  A  comes  out  an 
exact  number  of  times  U.  The  ratio  is  then  a  whole  number, 
and  we  say  that  U  exactly  measures  A,  or  that  ^  is  a 
multiple  of  U, 

Case  IL  We  may  find  that,  at  the  end,  the  measure  does 
not  come  out  exact,  but  a  piece  of  A  less  than  U  is  left  over. 
Or,  A  may  itself  be  less  than  U.  We  must  then  find  what 
fraction  of  U  the  piece  left  over  is  equal  to.  This  is  done  by 
dividing  U  up  into  such  a  number  of  equal  parts  that  one  of 
these  parts  shall  exactly  measure  A  or  the  piece  of  A  which  is 
left  over.  The  ratio  will  then  be  a  fraction  of  which  the  num- 
ber of  parts  into  which  U  is  divided  will  be  the  denominator, 
and  the  number  of  these  parts  in  A  the  numerator. 

Example.    If  we  find  that  i         i     ;         i  _.  jj 

by  dividing  U  into  7  parts,  4  of  p^—^-— ^-. 

these  parts  will  exactly  make  A,  \     \     \     \     \  =  ^ 

then  ^  =  4  of  U,  and  we  have  for  the  ratio  of  A  to  Z7, 

A:  U=^. 

If  we  find  that  A  contains  U  3  times,  and  that  there  is 
then  a  piece  equal  to  4-  of  Z7  left  over,  we  have 


130  RATIO. 

The  3  U^B,  are  equal  to  ^i}-  of  Z7,  so  that  we  may  also  say 

A  =  ^j-oiU,     or    A:U=^. 

which  is  simply  the  result  of  reducing  the  ratio  3f  to  an  im- 
proper fraction. 

In  general,  if  we  find  that  by  dividing  U  into  n  parts,  A 
will  be  exactly  m  of  these  parts,  then 

A  :  IJ  =  —, 

n 

whether  m  is  greater  or  less  than  n. 

When  the  magnitude  of  A  measured  by  U  can  be  exactly 
expressed  by  a  vulgar  fraction,  A  and  U  are  said  to  be  com- 
mensurable. 

Case  III.  It  may  happen  that  there  is  no  number  or  frac- 
tion which  will  exactly  express  the  ratio  of  the  two  magnitudes. 
The  latter  are  then  said  to  be  incommensurable. 

150.  Theorem.  The  ratio  of  two  incommensurable 
magnitudes  may  always  be  expressed  as  near  the  true 
value  as  we  please  by  means  of  a  fraction,  if  we  only 
make  the  denominator  large  enough. 

Examples.  Let  us  divide  the  unit  of  measur-e  into  20 
parts,  and  suppose  that  the  antecedent  contains  more  than  28 
but  less  than  29  of  these  parts.  Then,  by  supposing  it  to  con- 
tain 28  parts,  the  limit  of  error  will  be  one  part,  or  ^V  of  the 
standard  unit. 

In  general,  if  we  wish  to  express  the  ratio  within  1  n^^  of 
the  unit,  we  can  certainly  do  it  by  dividing  the  unit  into  n  or 
more  parts,  or  by  taking  as  the  denominator  of  the  fraction  a 
number  not  less  than  n, 

Ilhcstration  hy  Decimal  Fractions,  The  square  root  of  2 
cannot  be  rigorously  expressed  as  a  vulgar  or  decimal  fraction. 
But,  if  we  suppose 

a/2  =  1.4      =  \^,    the  error  will  be  <  yV  ; 
a/2  =  1.41    =\U,       "  "      <T*o; 

a/2  =  1.414  =  iiJi,     "  "       <y^. 

etc.        etc.  etc.  etc. 


NATURE   OF  A    RATIO.  131 

Since  the  decimals  may  be  continued  without  end,  the 
square  root  of  2  can  be  expressed  as  a  decimal  fraction  with  an 
error  less  than  any  assignable  quantity.  This  general  fact  is 
expressed  by  saying : 

The  limit  of  the  error  which  we  make  by  representing 
an  incommensurable  ratio  as  a  fraction  is  zero. 

151 .  Ratio  as  a  Quotient.  From  Case  II  and  the  explana- 
tions which  precede  it  we  see  that  when  we  say 

we  mean  the  same  thing  as  if  we  had  said, 

^  is  f  of  U,    or    A  =L^  U. 

If  A  and  U  are  numbers,  we  may  divide  both  sides  of  this 
equation  by  U,  and  obtain, 

A_^ 

U  ~  l' 
We  therefore  conclude  that  when  A  and  U  are  numbers, 

A  '  TT  —  —. 

That  is,  ^  .  u  -  ^ 

Theorem.  The  ratio  of  two  numbers  is  equal  to  the 
quotient  obtained  by  dividing  the  antecedent  term  by 
the  consequent. 

In  the  case  of  magnitudes,  the  relation  of  a  ratio  to  a  quo- 
tient may  be  shown  thus  : 

Let  us  have  two  magnitudes  M  and  F,  such  that  M  is 
4  times  V.     Then  we  may  write  the  relation, 

Jf=4F. 

Dividing  by  4,  we  have 

4   ~ 

Since  F  is  not  a  number,  we  cannot,  strictly  speaking, 
multiply  or  divide  by  it.  But  we  may  take  the  ratio  of  M  to 
F  without  regard  to  number,  and  thus  find, 

M  \  F  =  4. 


132  RATIO. 

Eem.  The  theory  of  ratios  the  terms  of  which  are  magni- 
tudes and  not  numbers,  is  treated  in  Geometry. 

In  Algebra  we  consider  the  ratios  of  numbers,  or  of  mag- 
nitudes represented  by  numbers. 

153.  Def.  If  we  interchange  the  terms  of  a  ratio, 
the  result  is  called  the  Inverse  ratio. 

That  is,    U:  A  is  the  inverse  oi  A  :  U, 

If  U  ',  A=-, 

n 

then  U  =  —A, 

n 

Ttl. 

and  we  have,  by  dividing  by  — , 

m 

or  A  \  U  —  — 

m 

Because  —  is  the  reciprocal  of  — ,  we  conclude : 

Theorem,  The  inverse  ratio  is  the  reciprocal  of  the 
direct  ratio. 

Properties  of  Ratios, 

153.  Theorem  I.  If  both  terms  of  a  ratio  be  multi- 
plied by  the  same  factor  or  divided  by  the  same  divisor, 
the  ratio  is  not  altered. 

Proof.     Eatio  of  ^  to  ^  =  5  :  ^  =  2* 

If  7n  be  the  factor,  then 

Eatio  of  mB  to  mA  ==:  inB  :  tyiA  =  — ~,  =  -t^ 

niA       A 

the  same  as  the  ratio  of  ^  to  ^. 

154.  Theorem  IL  If  both  terms  of  a  ratio  be  in- 
creased by  the  same  quantity,  the  ratio  will  be  increased 


PROPORTION.  133 

if  it  is  less  than  1,  and  diminished  if  it  is  greater  than  1 ; 
that  is,  it  will  be  brought  nearer  to  unity. 

Example.  Let  the  original  ratio  be  2  :  5  =  |.  If  we  repeatedly  add 
1  to  both  numerator  and  denominator  of  the  fraction,  we  shall  have  the 
series  of  fractions, 

2       3       4       5       pf/» 

each  of  which  is  greater  than  the  preceding,  because 

*  «-|  =  A;  whence,  f  >  |. 

f  -  S-  =  A- ;  whence,  f  >  f . 

I  -  f  ==  A  ;  whence,  |  >  ^ 

etc.  etc. 

General  Proof,  Let  «  :  5  be  the  original  ratio,  and  let 
both  terms  be  increased  by  the  quantity  u,  making  the  new 
ratio  a-\-u  :  &  +  u.     The  new  ratio  mmus  the  old  one  will  be 

{h  —  a)u 

If  h  is  greater  than  «,  this  quantity  will  be  positive,  show- 
ing that  the  ratio  is  increased  by  adding  u.  If  ^  is  less  than  a^ 
the  quantity  will  be  negative,  showing  that  the  ratio  is  dimin- 
ished by  adding  u. 


CHAPTER    II. 

PROPORTION. 

155.  Def.    Proportion  is  an  equality  of  two  or 
more  ratios. 

Since  each  ratio  has  two  terms,  a  proportion  must  have  at 
least  four  terms. 

•    Def.    The  terms  which  enter  into  two  equal  ratios 
are  called  Terms  of  the  proportion. 

\i  a\h  be  one  of  the  ratios,  and  p  \  q  the  other,  the  pro- 
poiiion  will  be, 

a:h^p\q.   •  (1) 


134  PROPORTION, 

A  proportion  is  sometimes  written, 

a  \  h    : :   p  :  q^ 
which  is  read,  "  As  <^  is  to  &  so  is  p  to  g."     The  first  form  is  to  be  pre- 
ferred, because  no  other  sign  than  that  of  equality  is  necessary,  but  the 
equation  may  be  read,  "  As  a  is  to  &  so  is  ^  to  g,"  whenever  that  expres- 
sion is  the  clearer. 

Def.  The  first  and  fourth  terms  of  a  proportion  are 
called  the  Extremes,  the  second  and  third  are  called 
the  Means. 

Theorems  of  Proportion. 

156.  Theorem  I.  In  a  proportion  the  product  of 
the  extremes  is  equal  to  the  product  of  the  means. 

Proof.  Let  us  write  the  ratios  in  the  proportion  (1)  in  the 
form  of  fractions.     It  will  give  the  equation. 

Multiplying  both  sides  of  this  equation  by  hq,  we  shall  have 
aq  =  b]),  (3) 

Cor.  If  there  are  two  unknown  terms  in  a  propor- 
tion, they  may  be  expressed  by  a  single  unknown 
symbol. 

Example.    If  it  be  required  that  one  quantity  shall  be  to 

another  as  jj  to  q,  we  may  call  the  first  px  and  the  second  qx, 

because 

px  \  qx  z=i  p  :  q  (identically). 

15*7.  Theorem  IL  If  the  means  in  a  proportion  be 
Interchanged,  the  proportion  will  still  be  true. 

Proof,  Divide  the  equation  (3)  by  pq.  We  shall  then 
have,  instead  of  the  proportion  (1), 

V'^  q 
or  a  :  p  =1  h  \  q. 


PROPORTION.  135 

Def.  The  proportion  in  which  the  means  are  inter- 
changed is  called  the  Alternate  of  the  original  pro- 
portion. 

The  following  examples  of  alternate  proportions  should  be  studied, 
and  the  truth  of  the  equations  proved  by  calculation  : 

1  :  2  rn    4:8;       alternate,     1:4    =2:8. 
2:3=6:9;  "  2:6=3:9. 

5  :  2  =  25  :  10 ;  "  5  :  25  =  2  :  10. 

158.  Theorem  IIL  If,  in  a  proportion,  we  increase 
or  diminish  each  antecedent  by  its  consequent,  or  each 
consequent  by  its  own  antecedent,  the  proportion  will 
still  be  true. 

Example.    In  the  proportion, 

5  :  2  =  25  :  10, 

the  antecedents  are  5  and  25,  the  consequents  2  and  10  (§  148).     Increasing 
each  antecedent  by  its  own  consequent,  the  proportion  will  be 
5  +  2  :  2  =  25  +  10  :  10,        or        7  :  2  =  35  :  10. 

Diminishing  each  antecedent  by  its  consequent,  the  proportion  will 
become, 

5  -  2  :  2  =  25  - 10  :  10,        or        3  :  2  =  15  :  10. 

Increasing  each  consequent  by  its  antecedent,  the  proportion  will  be 

5  :  2  +  5  ^  25  :  10  +  25,        or        5  :  7  =  25  :  35. 

These  equations  are  all  to  be  proved  numerically. 

General  Proof,     Let  us  put  the  proportion  in  the  form 

h        q  ^  ^ 

If  we  add  1  to  each  side  of  this  equation  and  reduce  each 
side,  it  will  give 

^+  ^ _  p  +  q 

~  I      -     ~q     ' 
that  is,  a  -^1)  :  h  =:  p  -]-  q  \  q,  (5) 

In  the  same  way,  by  subtracting  1  from  each  side,  it  will  be 
a  —  h  \  h  ^^  p  —  q  :  q.  (6) 


136  PROPORTION. 

If  we  invert  the  fractions  in  equation  (4),  the  latter  will 
become 

a       I) 

By  adding  or  subtracting  1  from  each  side  of  this  equation, 
and  then  again  inverting  the  terms  of  the  reduced  fractions, 
we  shall  find, 

Tlie  form  (5)  was  formerly  designated  as  formed  "by  composition," 
and  (6)  as  formed  "  by  division."  But  these  terms  are  now  useless,  be- 
cause all  the  above  forms  are  only  special  cases  of  a  more  general  one  to 
bL^  now  explained. 

159.  Theorem  IV,  If  four  quantities  form  the  pro- 
portion 

a  :  b  =  c  :  d^  {a) 

and  if  m,  n^  p^  and  q  be  any  multipliers  whatever,  we 
shall  have 

ma  +  nb  :  pa  -\-  qb  —  mc  -\-  nd  \  pc-\-  qd. 
Proof,     The  proportion  {a)  gives  the  equation, 


a 
h  ~ 

c 
d' 

Multiplyii 
member, 

ig  this  equation 

by  -   and 

adding   1   to 

each 

qh 

qd 

Keducing 

each  member  to 

a  fraction 

and 

inverting 

the 

terms. 

qh       _ 

qd 

pa  -^  qh        pc  -{-  qd 

Dividing  both  members  by  q, 

h__ d___ 

pa  -\-  qh'~  pc  +  qd 

The  original  proportion  {a)  also  gives,  by  inversion, 


(7) 


PROPORTION,  137 

from^which  we  obtain,  by  multiplying  by  -,  adding  1,  etc., 

qh  -\-  pa  _  qcl  +  pc 
pa       ~       pc 

^_  ^  __J_ /g) 

pa  +  qh       pc  -\-  qd 

(8)  xm  +  (7)  X  w  gives  the  equation, 

ma  -\-  nh  __  mc  +  7id 

pa  -{-  qh  '"  pc  -\-  qd  ^ 
or  ma  +  nh  :  pa  -\-  qb  :=  mc  +  nd  \  pc  +  qd,  (9) 

which  is  the  result  to  be  demonstrated. 

160.  Theorem  V.  If  each  term  of  a  proportion  be 
raised  to  the  same  power,  the  proportion  will  still 
subsist. 

Proof.     If  a  :  b  =  p  :  q, 

a       p 
or  T  =     y 

b        q' 

then,  by  multiplying  each  member  by  itself  repeatedly,  we 
shall  have 


Hence,  in 

l^"  q^' 
a^       p^ 
b^  ""  q^'' 

etc.     etc. 
general. 

a^  :  b"'  =z  p^  :  (p. 

Cor.    If 

a  :  b  =^  p  '  q^ 

then 

a^  \  a''  ±b^  —  pi^  \  p''  ±(f'  \ 

and 

a^±b^  \  If'  =  p^±q^  I  ^». 

Theorem  VI.  When  three  terms  of  a  proportion 
are  given,  the  fourth  can  always  be  found  from  the 
theorem  that  the  product  of  the  means  is  equal  to  that 
of  the  extremes. 


138  PROPORTION, 

We  have  shown  that  whenever 

a  :  h  =z  2^  :  q, 
then  aq  ==  hp. 

Considering  the  different  terms  in  succession  as  unknown 
quantities,  we  find, 


a 

J. 

~  q' 

I 

_aq 
-  P' 

P 

aq 

Q 

~"   a 

Cor.  1.  If,  in  the  general  equation  of  the  first 
degree 

ax  +  l)y  =^  c?5 

the  term  c  vanishes,  the  equation  determines  the  ratio 
of  the  unknown  quantities. 

Proof.     If  ax  +  by  =  0, 

then  ax  =  —  %, 

X  b 

and  -  = , 

y  a 

or  X  :  y  =z  —  b  :  a. 

Cor.  2.  Conversely,  if  the  ratio  of  two  unknown 
quantities  is  given,  the  relation  between  them  may  be 
expressed  by  an  equation  of  the  first  degree. 

The  Mean  Proportional. 

161.  Def.  When  the  middle  terms  of  a  proportion 
are  equal,  either  of  them  is  called  the  Mean  Propor- 
tional between  the  extremes. 

Tlie  fact  that  b  is  the  mean  proportional  between  a  and  c 
is  expressed  in  the  form, 

a  :  b  =^  b  :  c. 


PROPORTION,  139 

Theorem  I  then  gives,  b'^  =z  ae. 

Extracting  the  square  root  of  both  members,  we  have 

b  =  Vc^c, 
Hence, 

Theorem  VIL  The  mean  proportional  of  two  quan* 
tities  is  equal  to  the  square  root  of  their  product. 

Multiple  Proportions. 

163.  We  may  have  any  number  of  ratios  equal  to  each 
other,  as 

a  \  b  ^=i  c  :  d  =:  e  :  f,  etc. 

6  :  4  =  9  :  6  =  3  :  2  =  21  :  14.  (a) 

Such  proportions  are  sometimes  written  in  the  form 

6  :  9  :  3  :  21  =  4  :  6  :  2  :  14.  {b) 

In  the  form  {b)  tlie  antecedents  are  all  written  on  one  side 
of  the  equation,  and  the  consequents  on  the  other.  Any  two 
numbers  on  one  side  then  have  the  same  ratio  as  the  cor- 
responding two  on  the  other,  and  the  proportions  expressed  by 
this  equality  of  ratios  are  the  alternates  of  the  original  propor- 
tions {a).     For  instance,  in  the  proportion  {b)  we  have, 

6:9    =4:6,  which  is  the  alternate  of  6  :  4  =    9:6. 

6:3=4:2,       ''         "  "  6:4=3:2. 

6  :  21  =  4  :  14,     "         "  "  6  :  4  =  21  :  14. 

9  :  21  =  6  :  14,    "         "  "  9  :  6  =  21  :  14. 

163.  A  multiple  proportion  may  also  be  expressed  by  a 
number  of  equations  equal  to  that  of  the  ratios.     Since 

a  \  b  :=  c  \  d  z=z  e  :  f,  etc., 

let  us  call  r  the  common  value  of  these  ratios,  so  that 

a  c  , 

-^  =  r,        -^  =  r,    etc. 

Then  a  =  rb, 

c  =  rd,  (c) 

c  =  rf, 


140  PBOPORTION. 

will  express  the  same  relations  between  the  quantities  a,  2>,  c, 
dy  e,  f,  etc.,  that  is  expressed  by 

a  I  i  ^=^  c  :  d  =z  e  :  f,  etc.,  {a) 

or                     a  :  c  :  e  :  etc.  =z  h  \  d  :  f  :  etc.  {h) 

It  will  be  seen  that  where  r  enters  in  the  form  (c)  there  is  one  more 
equation  than  in  the  first  form  {a).  [In  this  form  each  =  represents  an 
equation.]  This  is  because  the  additional  quantity  r  is  introduced,  by 
eliminating  which  we  diminish  the  number  of  equations  by  one,  as  in 
eliminating  an  unknown  quantity. 

164.  Theorem,  In  a  multiple  proportion,  the  sum 
of  any  number  of  the  antecedents  is  to  the  sum  of  the 
corresponding  consequents  as  any  one  antecedent  is  to 
its  consequent. 

Ex.    We  have  -  —  --=--  =  -—'   Then 
5      15      25      30 

2  +  6  +  10  +  12       30 


5  +  15  +  25  +  30      75' 
which  has  the  same  value  as  the  other  four  functions. 

General  Proof,     Let  A,  B,  C,  etc.,  be  the  antecedents,  and 
a,  I,  c,  etc.,  the  corresponding  consequents,  so  that 


B  : 

1=  0  \ 

:  c, 

etc. 

(1) 

ion 

ratio  A  : 

a, 

B: 

b, 

etc.. 

so 

that 

A 

=  ra, 

B 

=  rh, 

C 

=  re. 

etc. 

etc. 

Adding  these  equations,  we  have 

A  +  B  -{-  C+  etc.  =  r{a  +  b  +  c-^  etc.), 

.      A  +  B  -{-  C  -\-  efcc. 

a  -\- h  -\-  c  -\-  etc.  ' 

that  is,  the  ratio  A -\- B -\- C -\- eta,  :  a -{- d -{- c -\- etc.  is  equal  to 
r,  the  common  value  of  the  ratios  A  :  a,  B  :  b,  etc. 

PROBLEMS. 

I .  A  map  of  a  country  is  made  on  a  scale  of  5  miles  to 
3  inches. 


PBOPOETION.  141 

(1.)  What  will  be  the  length  of  8,  12,  17,  20,  33  miles  on 
the  map  ? 

(2.)  How  many  miles  will  be  represented  by  6,  8,  16,  20, 
29  inches  on  the  map  ? 

Rem.  1.  If  Xy  y,  z,  u,  v  be  the  required  spaces  on  the  map,  we  shall 
have 

5  :  3  ==  8  :  aj  =  12  :  y,  etc. 

If  «,  b,  c,  etc.,  be  the  required  number  of  miles,  we  shall  have 
S  :  6  =  6  :  a  =  S  :  b  rzilQ  :  c,  etc. 

Rem.  2.  When  there  are  several  ratios  compared,  as  in  this  problem, 
it  will  be  more  convenient  to  take  the  inverse  of  the  common  ratio,  and 
multiply  the  antecedent  of  each  following  ratio  by  it  to  obtain  the  conse- 
quent:     In  the  first  of  the  above  proportions  the  inverse  ratio  is  f ,  and 

X  =  iotS,    y  =  f  of  12,  etc. 

In  the  second,       a  =:  f  of  6,     &  =  f  of  8,  etc. 

2.  To  divide  a  given  quantity  A  into  three  parts  whicli 
shall  be  proportional  to  the  given  quantities  a,  h,  c,  that  is, 
into  the  parts  x,  y,  and  z,  such  that 

X  \  a  =1  y  :  t  =^  z  :  c, 
or  X  \  y  :  z  ^=^  a  '.  h  \  c. 

Solution.    By  Theorem  IV, 

X y z X  -\-  y  -\-  z  A 

a~h~c~~a+h-\-c'~'a-\-h  +  c 
Therefore, 

_         aA  _         ^^  _         ^^ 

^  "~  a  +  ^  +  c'     ^  ""  a  +  b  -\-  c'     ^  ~  a  -\- h  +  c 

3.  Divide  102  into  three  parts  which  shall  be  proportional 
to  the  numbers  2,  4,  11.- 

4.  Divide  1000  into  five  parts  which  shall  be  proportional 
to  the  numbers  1,  2,  3,  4,  5. 

5.  Find  two  fractions  whose  ratio  shall  be  that  of  a:b,  and 
whose  sum  shall  be  1. 

6.  What  two  numbers  are  those  whose  ratio  is  that  of  7  :  3 
and  whose  difference  is  24. 

7.  What  two  numbers  are  those  whose  ratio  is  m  :  n,  and 
whose  difference  is  unity  ? 

8.  Find  x  and  y  from  the  conditions, 

X  :  y  =  a  :  b, 
ax  —  by  =  a  -{-  b. 


142  PROPORTION. 

9.  Show  that  if        a  :  b  =  A  :  B, 

c  :  d=i  C  :  D, 
we  must  also  have      ac  :  bd  =  AC  :  BD, 

10.  Having  giveu  x  =  ay,  find  the  value  of  ^  "^  ^^. 

x-2y 


II.  Having  given 


find  the  value  of 


x  —  'Zy 

x  +  y 


'^  =  5, 


x  —  y 
12.  If  a  :  b  ^=  p  :  q, 

a^  2j3 

prove  a^  +  b^  : 7  ^^  p^  ■\-  q^  \  —^- — , 

a-\-  b      ^^'^      i^+S' 


13.  If 


a  -\-  b  -{-  c  -}-  d       a  —  b  -{-  c  —  d 


a  +  b  —  c  —  d       a  —  b  —  c  +  d^ 
show  that  a  \  b  =:  c  \  d. 

14.  A  year's  profits  were  divided  among  three  partners,  A, 
B,  and  C,  proportional  to  the  numbers  2,  3,  and  7.  If  0 
should  pay  B  $1256,  their  shares  would  be  equal.  What  was 
the  amount  divided  ? 

15.  In  a  first  year's  partnership  between  A  and  B,  A  had 
2  shares  and  B  had  5.  In  the  second  year,  A  had  3  and  B  had  4. 
In  the  second  year,  A's  profits  were  $3200  greater  and  B's  were 
$1700  greater  than  they  were  the  first.  What  was  each  year's 
profits  ? 

16.  In  a  poultry  yard  there  are  7  chickens  to  every  2  ducks, 
and  3  ducks  to  every  2  geese.  How  many  geese  were  there  to 
every  42  chickens  ? 

17.  A  drover  started  with  a  herd  containing  4  horses  to 
every  9  cattle.  He  sold  148  horses  and  108  cattle,  and  then 
had  1  horse  to  every  3  cattle.  How  many  horses  and  cattle 
had  he  at  first  ? 

18.  If  a  bowl  of  punch  contains  a  parts  of  water  and  b 
parts  of  wine,  what  is  the  ratio  of  the  wine  to  the  whole 
punch  ?  What  is  the  ratio  of  the  water?  What  are  the  sums 
of  these  ratios  ? 


PROPORTION.  143 

19.  One  ingot  consists  of  equal  parts  of  gold  and  silver, 
while  another  has  two  parts  of  gold  to  one  of  silver.  If  I 
combine  equal  weights  from  these  ingots,  what  proportion  of 
the  compound  will  be  gold  and  what  proportion  silver  ? 

20.  What  will  be  the  proportions  if,  in  the  preceding  prob- 
lem, I  combine  one  ounce  from  the  first  ingot  with  three  from 
the  second  ? 

21.  One  cask  contains  a  gallons  of  water  and  h  gallons  of 
alcohol,  while  another  contains  m  gallons  of  water  and  n  of 
alcohol.  If  I  draw  one  gallon  from  each  cask  and  mix  them, 
what  will  be  the  quantities  of  alcohol  and  water  ? 

22.  What  will  be  the  ratio  of  the  liquors  in  the  last  case,  if 
I  mix  two  parts  from  the  first  cask  with  one  from  the  second  ? 

23.  What  will  it  be  if  I  mix  p  parts  from  the  first  with  q 
parts  from  the  second  ? 

24.  A  goldsmith  has  two  ingots,  each  consisting  of  an  alloy 
of  gold  and  silver.  If  he  combines  two  parts  from  the  first 
ingot  with  one  from  the  second,  he  will  have  equal  parts  of 
gold  and  silver.  If  he  combines  one  part  from  the  first  with 
two  from  the  second,  he  will  have  3  parts  of  gold  to  5  of  silver. 
What  is  the  composition  of  each  ingot  ? 

Suggestion.    Call  r  the  ratio  of  the  weight  of  gold  in  the  first  ingot 
to  the  whole  weight  of  the  ingot ;  then  1  —  r  will  be  the  ratio  of  the  sil-  * 
ver  in  the  first  to  the  whole  weight  of  the  ingot.     See  the  following 
question. 

Note.  Problems  18-24  form  a  graduated  series,  introductory  to  the 
processes  of  Problem  24. 

25.  Point  out  the  mistake  which  would  be  made  if  the 
solution  of  the  preceding  problem  were  commenced  in  the  fol- 
lowing way  : 

If  the  first  ingot  contains  p  parts  of  gold  to  q  parts  of  silver,  and  the 
second  contains  r  parts  of  gold  to  8  of  silver,  then 

Two  parts  from  the  first  ingot  will  have  2p  of  gold  and  2q  of  silver. 

One  part  from  the  second  ingot  will  have  r  of  gold  and  «  of  silver. 

Therefore,  the  combination  will  contain  2p  +  r  parts  of  gold,  and 
^q-\-s  parts  of  silver. 

Show  also  that  if  we  subject  p,  q,  r,  and  s  to  the  condition 
p-V-q  =  r  +  s, 
the  process  would  be  correct. 

26.  Show  that  if  the  second  term  of  a  proportion  be  a 
mean  proportional  between  the  third  and  fourth,  the  third 
will  be  a  mean  proportional  between  the  first  and  second. 


BOOK   V. 
OF    POWERS    AND    ROOTS. 


CHAPTER     I. 

INVOLUTION. 


Case  L   Involution  of  Products  and  Quotients. 

165.  Def.  The  result  of  taking  a  quantity,  A^ 
n  times  as  a  factor  is  called  the  n*'*^  power  of  A^  and 
as  already  known  may  be  written  either 

^  A  J.,  etc.,  72,  times,    or    A^. 

Bef.  The  number  n  is  called  the  Index  of  the 
power. 

Bef,  Involution  is  the  operation  of  finding  the 
powers  of  algebraic  expressions. 

The  operation  of  involution  may  always  be  expressed  by 
the  application  of  the  proper  exponent^  the  expression  to  be 
involved  being  inclosed  in  parentheses. 

Example.    The  n^^  power  of  «  +  J  is  (aj  +  lY. 
The  iif'^  power  of  abo  is  {abcY, 

166.  Involution  of  Products.  The  n^^  power  of  the 
product  of  several  factors  a^  h^  c,  may  be  expressed  without 
exponents  as  follows : 

dbcahcahC)   etc., 

each  factor  being  repeated  n  times. 


INVOLUTION,  145 

Here  there  will  be  altogether  n  a's,  n  Vs,  and  n  (fs,  so 
that,  using  exponents,  the  whole  power  will  be  a'^b^c^  (§  6^,  67). 

Hence,  {abc)^  =  aP'hH^ 

That  is, 

Theorem.     The  power  of  a  product  is  equal  to  the 
product  of  the  powers  of  the  several  factors. 

167.  Involution  of  Quotients.    Applying  the  same  methods 
to  fractions,  we  find  that  the  n^^  power  of  -  is  —  •    For 

(xY'      XXX 

(-)  = ,  etc.,  n  times: 

V      yyy 

_  XXX,  etc.,  n  times  .^  .  ^ 
~"  yyy^  ^tc,  n  times  ^^  ' ' 
__  x^ 

EXERCISES. 

Express  the  cubes  of 

I.     ahc,  2.     —  •  3.     ahcK 

c 

mn  a  -\-  h  mn  {a  +  5) 

pq  a  —  h  '     pq  {a  —  b)' 

Express  the  n^^  powers  of  the  same  quantities,  the  quanti- 
ties between  parentheses  being  treated  as  single  symbols. 

Case  II.    Involution  of  Powers. 

168.  Problem.     It  is  required  to  raise  the  quantity  a^  to 
the  n^^  power. 

Solution.     The  n*^  power  of  a^  is,  by  definition, 

a'^  X  a^  X  a^y  etc.,  n  times. 

By  §  66,  the  exponents  of  a  are  all  to  be  added,  and  as  the 
exponent  m  is  repeated  n  times,  the  sum 

m  -f  m  -h  m  4-  etc.,  n  times, 
is  m7i.    Hence  the  result  is  a^^,  or,  in  the  language  of  Algebra, 

10 


146  IJS  VOLUTION, 


Hence, 

^eorem.  If  any  power  of  a  quantity  is  itself  to  be 
raised  to  a  power,  the  indices  of  the  powers  must  be 
multiplied  together. 

EXAM  P  LE  S. 

Note.  It  will  be  seen  that  this  theorem  coincides  with  that  of  Case  I 
when  any  of  the  factors  have  the  exponent  unity  understood. 

EXERCISES. 

Write  the  cubes  of  the  following  quantities: 
4.a 


I.     dxyl 

2. 

3- 

aP", 

4.     bx'. 

5. 

2ahn\ 

6. 

6a^ 
b 

Write  the  71^^  powers  oi 

7.     a. 

8. 

a^. 

9- 

aWc, 

10.     a'^a^. 

II. 

2p^q\ 

12. 

{a-\-b)  {c-^d). 

^3.    0'^  +  «/)(^- 

-y)- 

14.     7  {a  +  b  — 

c){a- 

-by. 

Ans,  7^ 

'{a-\-b- 

■  c)^  (a  —  b^P. 

a 

'5-  r 

16. 

a2 

17. 

x-\-y 
x-y 

10.           „  • 
xy^ 

^''''   ~xnfn     • 

ab  {c  -  df 
"9-      (^  _  ^)  c3  • 

Reduce : 

20.     {^ab^n^f. 

21. 

(—  ^mnx^f. 

22.     2a{—3b^m7i^)K 

23. 

{IpqVy. 

24.     {ab^y^ 

25- 

{2ah^)\ 

26.     {m^)\ 

Note  1.  If  the  student  find  any  of  these  exponential  expressions 
difficult  of  expression,  he  may  first  express  them  by  writing  each  quantity 
a  number  of  times  indicated  by  its  exponent. 

Note  3.  The  student  is  expected  to  treat  the  quantities  in  paren- 
theses as  single  symbols. 


INVOLUTION.  147 

Eem.  The  preceding  theorem  finds  a  practical  application 
when  it  is  necessary  to  raise  a  small  number  to  a  high  power. 
If,  for  example,  we  have  to  raise  2  to  the  30th  power,  we 
should,  without  this  theorem,  have  to  multiply  by  2  no  less 
thf».n  29  times.  But  we  may  also  proceed  thus : 
2^  =  4, 

2^  =  22.22   =  4-4        ^  16, 
28  =  24.2^   =16.16     =  256, 
216  :=  28.28  =  2562       ^  65536^ 
224  ^  216.28  =  216.256  =  16777216, 
230  ^  221.26  =  224.64    =  1073741824. 

Case  of  IVegative  Exponents, 

169.  The  preceding  theorem  may  be  applied  to  negative 
^^xponents.     By  the  definition  of  such  exponents, 

g  =  aP5-..  (1) 

Eaising  the  first  member  to  the  n^^  power,  we  have. 

This  is  the  same  result  we  should  get  by  applying  the 
theorem  to  the  second  member  of  (1),  and  proves  the  proposi- 
tion. 

EXERCISES. 

Express  the  6th  powers  of 

I.     alrK 
3.     a7np~K 

Eeduce : 

II.     {ab-^c-^)-^. 
13.     {x^y-i)-\ 

After  forming  the  expressions,  write  them  all  with  positive 
exponents,  in  the  form  of  fractions. 


2. 

a^h'\ 

4. 

a-^h-\ 

6. 

{x-^yY{x-\-z) 

Q 

{a  +  h)-^ 

0. 

{a  -  by 

10. 

{ah-^c-'^y. 

12. 

{m^n-J)-^, 

14. 

\a%^c-''Y. 

148  INVOLUTION. 

Algebraic  Signs  of  Powers. 

110.  Since  the  continued  product  of  any  number  of  posi- 
tive factors  is  positive,  all  the  powers  of  a  positive  quantity  are 
positive. 

By  §  26,  the  product  of  an  odd  number  of  negative  fac- 
tors is  negative,  and  the  product  of  an  even  number  is  positive. 
Hence, 

Theorem.  The  even  powers  of  negative  quantities 
are  positive,  and  the  odd  powers  are  negative. 

EXAM  PLES.     ' 

(—  of  =  a^ ;     (—  ay  =:  —a^;     (—  aY  =  a*,     etc. 

EXERCI  SES. 


Find  the  value  of 

I.     (-3)^. 

2.     1 

[—  3)'- 

3- 

4*. 

4.     (-5)^. 

5-     < 

[-  6)3. 

6. 

{-by. 

7.     {-a -by. 

8.     ( 

[—mny. 

.  9- 

i-pqy. 

lO.     (— «)2». 

II. 

[-  5)2»+>. 

12. 

(_a_J)2«-i 

13.     (-1)^. 

14. 

[_i)2n+t. 

15- 

(-  l)^->. 

Case  III.  Involution  of  Binomials— the  Bino- 
mial  Theorem. 

1*71.  It  is  required  to  find  the  n^^  power  of  a  hinomial. 
1.  Let  a  +  ^  be  the  binomial ;  its  n^^  power  may  be  written 

Let  us  now  transform  this  expression  by  dividing  it  by  «^, 
and  then  multiplying  it  by  «%  which  will  reduce  it  to  its  orig- 
inal value.     We  have  (§  167), 

Multiplying  this  last  expression  by  V,  by  writing  this 
power  outside  the  parentheses,  it  becomes 


«^^ 


('  -  'J.  « 


INVOLUTION.  '  151 

preceding  one,  until  we  come  to  the  s^^  or  last,  which  will  be 

71  —  8  -{-  1. 

Such  a  ijroduct  is  written, 

n(?i  —  l){n  —  2) (n  —  s  +  l). 

The  dots  stand  for  any  number  of  omitted  factors,  because 
s  may  be  any  number.  We  have  written  4  of  the  s  factors,  so 
that  s  —  4  are  left  to  be  represented  by  the  dots. 

The  denominator  of  the  fraction  is  the  product  of  the  s 

factors, 

1.2.3....  5, 

each  factor  being  greater  by  1  than  the  preceding  one,  and  the 
dots  standing  for  any  number  of  omitted  factors,  according  to 
the  value  of  s.     Thus,  the  s^^  coefficient  in  the  7i^^  line  will  be 

n{n-l)(n-2) (n  -  s  +  1) 

1.2.3 s  "  ^"^^ 

If  5  is  greater  than  ^n,  the  last  factors  will  cancel  some  of 
the  preceding  ones,  so  that  as  s  increases  from  ^n  to  n,  the 
values  of  the  preceding  coefficients  will  be  repeated  in  the 
reverse  order.  Thus,  suppose  n  =  6,  Then,  by  cancelling 
common  factors, 

= ^J = -• 

1.2.3.4.5.6  ~  ^' 

If  we  should  add  one  more  factor  to  the  nul^erator,  it 
would  be  0,  and  the  whole  coefficient  would  be  0. 

The  conclusion  we  have  reached  is  embodied  in  the  follow- 
ing equation,  which  should  be  perfectly  memorized : 

/-I    ,     \*,        -I    .          ,  n  hi  —  1)   ^      ii(n  —  l)  (n  —2)   . 
(1  +  x)n  =  1  +  iix  4-  --^372  —  ^  +  -^ iT^ ^ 

^  1.2.3.4  ^  ^^***  ^-^'^ 


6.-5.4.3 

1-2.-3.4 

6.5.4.3.2 

1.2.3.4.5 

6.5.4.3.2.1 

152  INVOLUTION, 


EXERCISES. 

1.  Compute  from  the  formula  {d)  all  the  binomial  coeffi- 
cients for  ^  =:  6,  and  from  them  express  the  development  of 
(1  +  xf. 

2.  Do  the  same  thing  for  n  =  8,  and  for  n  =  10. 

173.    To  find  the  development  of  (a  -f-  b)^,  we  replace  x 

by  - ,  and  then  multiply  each  term  by  a^, 
(t 

[See  equations  (1)  and  (2).]     We  thus  have 

^7    i/ll     1^ 

{a  +  hY  z=i  oP'  ^  na^'-'^h  +     V      -^  a'^-W  +  etc.  to  l^ 

The  terms  of  the  development  are  subject  to  the  following 
rules : 

I.  The  exponents  of  b,  or  the  seeoncl  term  of  the  hiiio- 
mial,  are  0,  1,  2,  etc.,  to  n. 

Because  W  is  simply  1,  a"  is  the  same  as  W^V^. 

II.  The  sum  of  the  exponents  of  a  and  b  is  n  in  each 
term.    Hence  the  exponents  of  a  are 

n,     n  ~  1,     n  —  2,    etc.,  to     0. 

III.  The  coefficient  of  the  first  term  is  unity,  and  of 
the  second  n,  the  index  of  the  power.  Each  following 
coefficient  may  he  found  from;  the  next  preceding  one  hy 
multiplying  by  the  successive  factors, 

n  —  1  n  —  2  n  —  d 

""2~^         —3—,         — J-.    etc. 

IV.  //  b  or  a  is  negative,  the  sign  of  its  odd  powers 
will  he  changed,  hut  that  of  its  even  powers  will  remain 
the  same. 


(Compare  §  170.)     Hence, 
{a  —  hY  —  a^  —  na^-^b 
the  terms  being  alternately  positive  and  negative. 


{a  -  hY  =:a^-  nan-^b  +  ^^.-Jil  «7i-2^  _  etc., 


('-r^ 


INVOLUTION.  153 

EXERCISE  S  — Continued. 

3.  Compute  all  the  terms  of  {a  +  by,  using  the  binomial 
coefficients. 

4.  What  is  the  coefficient  of  W  in  the  development  of 
{a  +  ^)^«. 

5.  What  are  the  first  foui  terms  in  the  development  of 
{2am  +  3^)8. 

6.  What  are  the  first  three  terms  in  the  development  of 

Vl8 

?    What  are  the  last  two  terms  ? 

7.  What  are  the  first  three  and  the  last  three  terms  of 

8.  What  is  the  development  of  (a  +  -)  • 

9.  What  are  the  first  four  terms  in  the  development  of  the 
following  binomials : 

(1  +  a^)n ;  (1  +  2x^Y ;  (1  -  "^^^Y ; 

10.  What  are  the  sum  and  difference  of  the  two  develop 
ments,  (1  +  xf  and  (1  —  a;)^? 

Case  IV.    Square  of  a  JPolynoinial. 

173.    1.  Square  of  any  Polynomial    Let 
a-}-b  +  c-{-d  +  etc., 
be  any  polynomial.     We  may  form  its  square  thus : 

a-{-b-i-c  +  d-{-  etc. 

a  +  b-\-c-\-d-{-  etc^^ 

a^  -\-  ab  -\-  ac  -{-  ad  +  etc. 

ab  -{-b^  -{-be  -hbd  +  etc. 

ac  -}-  be  -{-  €^  -}-  cd  -^  etc. 

ad  4-  bd  -\-  cd  -{-  d^  +  etc. 

a^  -j-b^  -\-  c^  +  ^^  +  etc. 
+  2ab  +  2ac  +  2ad  +  etc. 

+  2bc  +  2bd  +  etc.  -f-  2cd  +  etc. 


164  INVOLUTION, 

We  thus  reach  the  following  conclusion : 

Theorem,  The  square  of  a  polynomial  is  eqnal  to 
the  sum  of  the  squares  of  all  its  terms  plus  twice  the 
product  of  every  two  terms. 

2.  Square  of  an  Entire  Function,  Sometimes  we  wish  to 
arrange  the  polynomial  and  its  square  as  an  entire  function  of 
some  quantity,  for  example,  of  x. 

Let  us  form  the  square  oi  a  -\-1jx  -{-  cx^  +  d^  +  etc. 

a  -\-  dx  ^  cx*^  -\-  dx^  +  etc. 
a  -\- hx  -^  cx^  -\-  dx^  -f  etc. 


«2  -f  adx  +  acx^  +  adx^  +  etc. 

abx  -f-  ¥x^  +  hcx^  -\-  bdx^  +  etc. 

acx^  -h  bcx^  +   c^x^  +  etc. 

ada^  -\-  hdx^  +  etc. 

a^  +  2al)x  +  {2ac  +  b^)  x^  -^-J^ad  +  2I)c)  x^  +  etc. 

We  see  that : 

The  coefficient  of  x^  is  ac  -\-  hh  -{-  ca, 

"  "  "  c^  is  ad-^lc  -^  cb  -\-  da, 

"  "  "  ^  is  ae  -\-  M  -{-  gg -\-  dh  -\-  ea, 

etc.  etc. 

The  law  of  the  products  ae,  hd,  gg,  etc.,  is  that  the  first 
factor  of  each  product  is  composed  successively  of  all  the  co- 
efficients in  regular  order  up  to  that  of  the  power  of  x  to  which 
the  coefficient  belongs,  while  the  second  factor  is  composed 
successively  of  the  same  coefficients  in  reverse  order. 

EXERCISES. 

Form  the  squares  of 

I.     1  +  2:i;  +  ^x\       ■  2,     1  -\-  2x  +  3x^  +  4^. 

3.  l-\-2x-{-  3x^  +  4:0^  +  5a;5. 

4.  1  +  2x  +  Sx^  +  ^x^  +  5a;5  +  (jx^ 

5.  1  —  2x  +  dx'^  —  4:X^  6.     a—b  -\-  c  —  d, 

a  0 


evolution:  155 

CHAPTER     11. 

EVOLUTION    AND    FRACTIONAL    EXPONENTS. 

174.  Def.  The  it*'*  Root  of  a  quantity  q  is  snch  a 
number  as,  being  raised  to  the  ii}^  power,  will  produce  q. 

When  n  =  2^  the  root  is  called  the  Square  Root. 

When  72,  =  3,  the  root  is  called  the  Cube  Root. 

'   Examples.    3  is  the  4th  root  of  81,  because 

3.3.3.3  ■=  34  =  81. 

As  the  student  already  knows,  we  use  the  notation, 

71^^  root  oiq  z=z  ^^. 

There  is  another  way  of  expressing  roots  which  we  shall 
now  describe. 

175.  Division  of  Exponents.  Let  us  extract  the  square 
root  of  a^.  We  must  find  sach  a  quantity  as,  being  multiplied 
by  itself,  will  produce  a^.  It  is  evident  that  the  required  quan- 
tity is  a^,  because,  by  the  rule  for  multiplication  (§§  ^Q,  166), 

■a^  X  a^  =  a', 

n 

The  square  root  of  a'^  will  he  a^,  because 

n  n  IL^Vl 

a^  X  0^  —  a^  ^  =  a^. 

n 

In  the  same  way,  the  cube  root  of  a^  is  a^,  because 

n  n^  n^ 

0^  X  a^  X  a^  =  a^. 
The  following  theorem  will  now  be  evident : 

Theorem.  The  square  root  of  a  power  may  be  ex- 
pressed by  dividing  its  exponent  by  2,  the  cube  root  by 
dividing  it  by  3,  and  the  n^^  root  by  dividing  it  by  n, 

176.  Fractional  Exponents.  Considering  only  the  origi- 
nal definition  of  exponents,  such  an  expression  as  n^  would 


156  EVOLUTION. 

have  no  meaning,  because  we  cannot  write  a  1^  times.  But 
by  what  has  just  been  said,  we  see  that  cfi  may  be  interpreted 
to  mean  the  square  root  of  a^,  because 

3  3 

a'^  X  a'^  —  a\ 
Hence, 

A  fractional  exponent  indicates  the  extraction  of  a 
root.  If  the  denominator  is  2,  a  square  root  is  indi- 
cated ;  if  3,  a  cube  root ;  if  n^  an  n^^^  root. 

A  fractional  exponent  has  therefore  the  same  meaning  as 
the  radical  sign  \/,  and  may  be  used  in  place  of  it. 

EXERCISES. 

Express  the  following  roots  by  exponents  only  : 

I.     "s/m,  2.     ^/(w^  +  7i).  3.     a/(«  +  ^)^. 

4.     ^/{a  +  hY.      5.     ^/mK  6.     ^^^ 

7.     \/a\  8.     ^/{a-\-bY,  9.     ^/{a  +  h)^. 

177.  Since  the  even  powers  of  negative  quantities 
are  positive,  it  follows  that  an  even  root  of  a  positive 
quantity  may  be  either  positive  or  negative. 

This  is  expressed  by  the  double  sign  ± . 

EXERCISES. 

Express  the  square  roots  and  also  the  cube  roots  and  the 
nf'^  roots  of  the  following: 

I.     (a  +  If.  2.     {a  +  hf.  3.     a-^h 

4.     {x-\-yf.  5.     (^  +  #.  6.     (:x  +  y)K 

178.  If  the  quantity  of  which  the  root  is  to  be  ex- 
tracted is  a  product  of  several  factors,  we  extract  the 
root  of  each  factor,  and  take  the  product  of  these  roots. 

Example.     The  n^^  root  of  am^p  is  a^ni^p^,  because 
{Jirn^p^Y  =  arri^p,  by  §§  168  and  176. 

If  the  quantity  is  a  fraction,  we  extract  the  root  of 
both  members. 


EVOLVTIOJSr. 


157 


Proof, 


(§§167,168.) 


Because  ~7  taken  7i  times  as  a  factor  makes  t  9  therefore, 
^n  0 

ct 
by  definition,  it  is  the  n*^  root  of  t- 

EXERCISES. 

Express  the  square  roots  of 

I.         4:X\  2.  -— 

Express  the  cube  roots  of 
4.     27-64.  5.     27«3. 

7.     ab^(^d\ 


6.     64.27«35«' 


8a^ 


125a;^^ 
Express  the  ^^^^  roots  of 
9.     7.  10.     4.7. 


12. 


6a^b^^, 


II.     4.7.10. 

6«2|^^ 


14. 


15- 


/^7W+l  ^/J  >^— 2 


C^cln 


dTnn  J^ 

16.  35^  a-'^""  (a  +  Z>)4'^  (r?;  —  ^jY  4^  (5  —  c  +  cZ)-^. 
Eeduce  to  exponential  expressions : 

17.  v^^r=^.  18.     ^^^^. 

19.      '^^^O^. 

+  bY 


20. 


21 


m/(a 


bY 


Powers  of  Expressions  with  Fractional  Expo- 
nents. 

179.  Theorem.    The  p^^  power  of  the  n^^  root  ig 
equal  to  the  n^^  root  of  the  p^^  power. 


158  FRACTIONAL    EXPONENTS. 

Ill  algebraic  language, 

or  (fl^-)"  —  (a^)", 

Example.  (v^s)^  =:  2^  ==  4, 

or,  in  words,  the  square  of  the  cube  root  of  8  (that  is,  the 
square  of  2)  is  the  cube  root  of  the  square  of  8  (that  is,  of  G4). 

Gejieral  Proof.     Let  us  put  :^;  =  the  n^^  root  of  a,  so  that 

x^  =  a.  (1) 

The  p^^  power  of  this  root  x  will  then  be  xp.  (2) 

Eaising  both  sides  of  the  equation  (1)  to  the  2^^^  power,  we 
have 

af'P  =  aP  =  pi^  power  of  a. 

The  71^^  root  of  the  first  member  is  found  by  dividing  the 
exponent  by  n,  which  gives 

n*^  root  ofp^^  power  =  x^, 

the  same  expression  (2)  just  found  for  the  p^^  power  of  the 
71^^  root. 

This  theorem  leads  to  the  following  conclusion : 

1.  The  expression  p 

a/" 

1 

may  mean  indifferently  the  p^^  power  of  a^,  or  the  ?ztli 
root  of  6i^,  these  quantities  being  identical. 

2.  The  pov^ers  of  expressions  having  fractional  ex- 
ponents may  be  formed  by  multiplying  the  exponents 
by  the  index  of  the  power. 

EXE  RCISES. 

Express  the  squares,  the  cubes,  and  the  7i^^  powers  of  the 
following  expressions : 

I.     a^.  2.     ak  3.     ai 

4.     n'^i  5.     ab^.  6.     ab^c^. 


IRRATIONAL    EXPRESSIONS.  159 


m  m 


P       Q 


7.     a^h^,  8.  a^h  v. 

m 

9.     {a  +  hf{a  —  b)~^.  10.  a-^h"". 

II.     a^h\  12.  ^    ^  ^-^  ^ 

Reduce  to  simple  products  and  fractions : 

^y  ^)  .  14.  {cfih^c'i)^, 

15.     {aH^y^.  16.  \a   ^1    9, 


(^:) 


'^-       \-A}    •  '^-       .^^    •    .-R- 


CHAPTER    III. 

REDUCTION    OF    IRRATIONAL    EXPRESSIONS. 


Definitions. 


180.  Def.  A  Rational  Expression  is  one  in  which 
the  symbols  are  only  added,  subtracted,  multiplied,  or 
divided. 

All  the  operations  we  have  hitherto  considered,  except  the  extraction 
of  roots,  have  led  to  rational  expressions. 

Def.  An  expression  which  involves  the  extraction 
Oi  a  root  is  called  Irrational. 

Example.     Irrational  expressions  are 

Va,        ^/{a  +  h),         a/27; 
or,  in  the  language  of  exponents, 

^4,      .     {a  +  ^)i,  27*. 

In  order  that  expressions  may  be  really  irrational. 


160  IRRATIONAL   EXPRESSIONS. 

they  must  be  Irreducible,  that  is,  incapable  of  being 
expressed  without  the  radical  sign. 

Example.     The  expressions 

^'cf^2ah  +  h^,         a/36,       ' 

are  not  properly  irrational,  because  they  are  equal  to  a  -\-  h 
and  6  respectively,  which  are  rational. 

Def.  A  Surd  is  the  root  which  enters  into  an 
irrational  expression. 

Example.  The  expression  a  +  hVx  is  irrational,  and  the 
surd  is  Vx. 

Def.  Irrational  terms  are  Similar  when  they  con- 
tain the  same  surds. 

Examples.  The  terms  ^30,  7^/30,  {x  +  y)  a/30,  are 
similar,  because  the  quantity  under  the  radical  sign  is  30  in 
each. 

The  terms  {a  +  h)  ^/x  -f-  y,  dVx  +  ^,  fnVx  +  y  are 
similar. 

Ag^gregation  of  Similar  Terms. 

181.  Irrational  terms  may  be  aggregated  by  the  rules  of 
§§  54—56,  the  surds  being  treated  as  if  they  were  single  sym- 
bols.    Hence : 

WTien  similar  irrational  terms  are  conneetecl  hy  the 
signs  -{-or  — ,  the  eoeffieients  of  the  similar  surds  may 
he  added,  and  the  surd  itself  affixed  to  their  sum. 

Example.    The  sum 

aV{x  +  y)  —  W{x  +  ij)  +  dV(x  +  y) 

may  be  transformed  into  (a  —  b  +  3)  V{x  +  y). 

EXERCISES. 

Keduce  the  following  expressions  to  the  smallest  number 
of  terms : 

I.     7a/3  —  5a/2  +  6a/3  +  7a^A/2. 


IRRATIONAL    EXPRESSIONS,  161 

2.    W{x  +  y)  +  W{x  -y)  +  2{a  +  b)  V{x  +  y) 

—  3{a  +  b)  V{x  —  y)' 

4.  {fl  +  y)  a/^2/  +  (a  —  Z>)  Viz^^. 

5.  Vx  {a  —  b)-\-(b  —  c)  ^/x  ■\-  {c  —  d)  Vx. 

6.  aVx  —  ^/x  +  "la^/x  —  {a  -\-  b)  Vx. 

7.  -  Vi?^  —  aVx  +  QV^  —  cVx  +  ^  v''^. 

8.  ^  "r     Va:  —  GcV^  —  "4~~  "^^  +  ^^' 

9.  -Vx  —  Vx+  {a  —  h)  Vx  +     ^^~    ^  Vi«^. 

10.  a/^  —  hVa  —  a/o;  H 4-"     '^^  ~  o  '^^" 

11.  jv^— V^H — ^— o — -yx, 

12.  4A/i?^  — •  Q  'V'^^  +  («^  —  ^)  Vi^. 

o 

Factoring  Surds. 

183.  Irrational  expressions  may  sometimes  be  transformed 
so  as  to  have  different  expressions  under  the  radical  sign,  by 
the  method  of  §  178,  applying  the  following  theorem : 

Theorem.  A  root  of  the  product  of  several  factors 
is  equal  to  the  product  of  their  roots. 

In  the  language  of  Algebra, 


^/abcd,  etc.  =  "s/a  ^/h  "s/c  \^d,  etc. 

Proof,     By  raising  the  members  of  this  equation  to  the 
n^^  power,  we  shall  get  the  same  result,  namely, 
a  X  I?  X  c  X  d,  etc. 

Example.     VsO  =  V^  a/S. 
11 


162  IRRATIONAL    EXPRESSIONS. 

EXERCISES. 

Prove  the  following  equations  by  computing  both  sides : 


a/4  V'49  =  V4:.49  =  Vl96. 

Pm/.      a/4:  a/49  r:^  2-7  =  14,  and  a/196  =:  14. 

V4  a/9  =  a/36. 

a/4  ^25  rr  a/4725. 

a/9  a/16  =  A/90L6. 

V25  Vse  =  a/25^. 

Express  with  a  single  surd  the  products : 

I.     V{a  +  d)  V{a  —  b). 

SoLUTioi;r.     V{a  +  b)  V(a  —b)  =  V{a  +  b){a  —  b) 

V7V6.  3.     V^Va. 

Va  V{a  +  y),  5.     Va  Vb  V{a  +  b). 

V{x  +  I)  V{x  -  1). 

V{x^  +  1)  V{x  +  1)  \/{x  -  1). 

l(a  +  ^)^  {a  -  b)ij. 

183.  If  we  can  separate  the  quantity  under  the 
radical  sign  into  two  factors,  one  of  which  is  a  perfect 
square,  we  may  extract  its  root  and  affix  the  surd  root 
of  the  remaining  factor  to  it. 

EXAMPLES. 


Va^b  =  Va^  Vb  —  aVb. 
Vab  Vac  =  Vcf^bc  =  aVbc, 


a/12  a/6  =:  a/72  =  a/36  a/2  =  6  a/2. 
V{^a^  +  8a^J  —  \^ah^  =  V^d^  {a  ^2f^~iac) 
—  2aV{ct  +  25  —  4ac). 
(x^—  Ax^y  -h  Ax}/)^  =  (a;  —  2^)  x^. 


IRRATIONAL    EXPRESSIONS.  163 

EXERCISES. 

Eeduce,  when  possible : 

I.     a/8.  2.  a/32. 

3.     a/128.  4.  a/3  a/27. 

5.     ^'ab  ^/'c~a  ^/~hc.  6.  a/2  a/72. 

7.     a/4  a/72.                  •  8.  a/(:z:  +  1)  a/(^  +  1). 

9.     a/175.  ro.  a/150c 

II.    a/108.  12.  a/^^^T^T+T). 
13.     a/(«^^  +  '^cbbx  +  ^2^). 
Here  the  quantity  under  the  radical  sign  is  equal  to 

In  questions  of  this  class,  the  beginner  is  apt  to  divide  an  expression 
like  ^a  +  6  H-  c  into  /y/(X  +  ^Jh  +  y'c,  which  is  wrong.  The  square 
root  of  the  sum  of  several  quantities  cannot  be  reduced  in  this  way. 


14.     \^dhj  At  ^CLy  •\-  ^y-  15.     ^/^mh  +  "^mz  -f  4;$;. 

Eeduce  and  add  the  following  surds : 
16.     4a/2  — 6a/8  +  10a/32.     17.     a/12  +  a/27 -f- a/75. 
18.      A/4a  — 2a/«.  19.     125^  —  45^  —  80*. 

20.     'v/81  —  A^192.  21.     {aWf^  +  (aSc^ji 

Mviltiplication  of  Irrational  Expressions. 

184.  Irrational  polynomials  may  be  multiplied  by  com- 
bining the  foregoing  principles  with  the  rule  of  §  78. 
The  following  are  the  forms  : 

To  multiply  a  +  h^/x  by  m  +  n^/y, 

a  (m  +  nVy)  =  (i^n  x  an^/y. 
hVx  {m  -f  nVy)  =  hm^/x  +  In^/xy, 
The  product  is  am  +  mWy  -}-  hnVx  +  In^/xy, 

EXERCI  SES. 

Perform   the   following   multiplications   and    reduce    the 
results  to  the  simplest  form  (compare  §  80) : 

I.     (2 +  3  a/5)  (5  — 3 a/2).         2.     (7  +  2 a/32)  (9  —  5 a/2). 


164  IRRATIONAL   EXPRESSIONS, 

3.  {a  +  ^/h)  {a  —  V^).        4.     {^/a-\-^/h-^Vc-{- \^d)\ 

5.  (m  +  n^)  (m  +  2/1*).         6.     («i  —  c^s)  (a^  -f-  a^), 

7.  (^  +  «-^)^.  8.     (a* -a-*)'. 

9.  [«  +  &\/(^  +  y)'\  [a  -  2>a/(^  +  y)l 

10.  [m  -|-  /^V(«5  -f  ^)]  [w  —  /^a/(«  —  h)\ 

11.  [a;  +  VCt'^  —  1)]  \x  —  V(^2  _  1)]. 

12.  [(^2  +  1)4  +  ^J  [(^H- l)i  -  ^]. 

Expressions  may  often  be  transformed  and  factored  by 
combining  the  foregoing  processes. 

Example.     To  factor  axi  +  bxi  +  cxi  +  dx~2,  we  notice 

that  1         X  a  5         i  ^       ; 

x-^  =  x'^x^,        x-^  =z  x-^x%     etc. 

so  that  the  expression  may  be  written, 

ax^xi  4-  bx^x^  +  cxxi  +  dxi  =  (ax^  +  hx^  -{-  ex  +  J)  x^, 

EXE  RCISES. 

Keduce  the  following  expressions  to  products : 

13.  2  4-V2.  14.     3^  +  2-3*. 


15.     {a  4-  Z>)2.  16.     V^  +  ay^  —  %3, 

17.     X  —  y  —  ^/x  —  y. 

Eeduce  to  the  lowest  terms : 


i».     — - 


2                           ^/a'^-h 

20. 
22. 

ax^  +  ho^ 
axi  —  bx^ 

a  —  X  +  Vcc  —  x 

Va^  -  b^ 

21,  *^. 

a  —  .T  —  Vet  —  X  a  -jr  b 

185.  Rationalizing  Fractions.  The  quotient  of 
two  surds  may  be  expressed  as  a  fraction  with  a  rational 
numerator  or  a  rational  denominator,  by  multiplying 
both  terms  by  the  proper  multiplier. 

Example.     Consider  the  fraction  -— • 


IRRATIONAL    EXPRESSIONS.  165 

Multiplying  both  terms   by   a/7,    the    fraction    becomes 

— — ,  and  has  the  rational  denominator  7. 

5 
Multiplying  by  y'S,  it  becomes  — — ,  and  has  the  rational 

numerator  5.  ^  ^^ 

The  numerator  or  denominator  may  also  be  made 
rational  when  they  both  consist  of  two  terms,  one  or 
both  of  which  are  irrational. 

Let  us  have  a  fraction  of  the*  form 

A  -^dVB 

P  +  Qy/R' 

in  which  the  letters  A,  D,  P,  Q,  and  R  stand  for  any  algebraic 
or  numerical  expressions  whatever.  If  we  multiply  both  nu- 
merator and  denominator  by  P  —  Q^ R,  the  denominator 
will  become 

p2  _  Q^n, 

The  numerator  will  become 

AP  +  PB^/B  -  A  QVR  -  DQ^BR. 
so  that  the  value  of  the  fraction  is 

AP  +  pdVb  -^  AQVR  -  dqVbr 
P'  -  Qm 

EXE  RCI  SES. 

Reduce  the  following  fractions  to  others  having  rational 
denominators : 

7a/3  2V18  .      5V24 

9V5  3\/6  2a/^ 


166                                PEMFEGT  SQUARES, 

a  +  2V(^  +  y)  2a/3  +  7a/5 

lO.       : •  II. 

a 


+  V(x  -\-y)                   *       V5  _  V3 
12.     V^  -  V(>^'  +  ^)  ^^^  ^ 

1  V^  +  «^  +  '\/x  —  a 

«i  +  (a  +  1)3  "s/x  4-  a  —  V^  —  a 

Perfect  Squares. 

186.  Def.  A  Perfect  Square  is  an  expression  of 
whicli  the  square  root  can -be  formed  without  any  surds, 
except  such  as  are  already  found  in  the  expression. 

Examples.  4/72^,  4^^  +  4^  +  1  are  perfect  squares^  be- 
cause their  square  roots  are  2m\  2a  +  1,  expressions  without 
the  radical  sign. 

The  expression  a  +  2Vcib  -\-  b,  of  which  the  root  is 

Vet  +  Vb, 

may  also  be  regarded  as  a  perfect  square,  because  the  surds 
Vcc  aad  \^b  are  in  the  product  2\/ab, 

Criterion  of  a  Perfect  Square,  The  question  whether  a 
trinomial  is  a  perfect  square  can  always  be  decided  by  compar- 
ing it  with  the  forms  of  §  80,  namely : 

a^  +  2ab  -^  b'^  =  {a  -^  b)% 
or  «2  _  2ab  +  ^2  —  (^  _  ^)2. 

We  see  that  to  be  a  perfect  square,  a  trinomial  must  fulfil 
the  following  conditions : 

(1.)  Two  of  its  three  terms  must  be  perfect  squares. 

(2.)  The  remaining  term  must  be  equal  to  twice  the 
product  of  the  square  roots  of  the  other  two  terms. 

When  these  conditions  are  fulfilled,  the  square  root 
of  the  trinomial  will  be  the  sum  or  difference  of  the 
square  roots  of  the  terms,  according  as  the  product  is 
positive  or  negative. 

The  root  may  have  either  sign,  because  the  squares  of  posi- 
tive and  negative  quantities  have  the  same  sign. 


COMPLETING    THE  SQUARE, 


167 


If  the  terms  which  are  perfect  squares  are  loth  negative, 
the  trinomial  will  be  the  negative  of  a  perfect  square. 


EXAMPLES. 

^/W~^ab~^^  =  a  -\-  b  or  —  (a  -{-  b). 
b  OY  b  ~a. 


^/^2  _2ab -\- b^  =  a- 
—  a^  +  2ab  —  b^=  —  {a  ■ 


of, 


EXERCISES. 

Find  which  of  the  following  expressions  are  perfect  squares, 
and  extract  their  square  roots : 


I. 

9  +  12  +  4. 

2. 

r?:^  +  4^  +  4. 

3. 

4^  _l_  2x^  +  h 

4. 

a2  ^ab  —  b\ 

5- 

4:a^^  +  12a^Z»^^+  9b'^^. 

6. 

a^  +  %ab  —  W. 

7. 

x^  —  ax^y  +  -  a^y\ 

8. 

aW  —  ^abccl  +  c^d^. 

9- 

771  +  2m^n^  +  n. 

10. 

a^  —  2ax  +  y\ 

II. 

a  +  4a4z>i  +  42>. 

12. 

f^  —  2  +  «-!. 

IS- 

25y +  9g2  —  ^Op\ 

14. 

ejim^n  _^  7^2  ^  9^^^4;i, 

IS- 

^9xY  +  9^2  _  42:r?/^. 

16. 

9m^^  —  2m>g  +  ^ 

To  Complete  the  Square. 

187.  If  one  term  of  a  binomial  is  a  perfect  square, 
such  a  term  can  always  be  added  to  the  binomial  that 
the  trinomial  thus  formed  shall  be  a  perfect  square. 

This  operation  is  called  Completing  the  Square. 

Proof,  Call  «^  the  root  of  the  term  which  is  a  perfect 
square,  which  term  we  suppose  the  Jirsi,  and  call  m  the  other 
term,  so  that  the  giyen  binomial  shall  be 

a^  +  m. 

Add  to  this  binomial  the  term  — ^ ,  and  it  will  become 


a^  +  7)7  + 


4^2 


168  COMPLETING    THE  SQUARE. 

This  is  a  perfect  square,  namely,  the  square  of 


m 

thatis,  a^  +  m  +  ^,  =  [a  +  -). 

Hence  the  following 

EuLE.     Add  to  the  hinomial  the  square  of  the  second 
term  divided  hy  four  times  the  first  tei^m. 

Example.     What  term  must  be  added  to  the  expression 
x^  —  4:ax  (1) 

to  make  it  a  perfect  square  ? 

The  rule  gives  for  the  term  to  be  added, 

4:X^ 

Therefore  the  required  perfect  square  is 

x^  —  ^ax  +  4r^2  ^  {-x  —  2af. 

We  may  now  transpose  4:0?,  so  that  the  left-hand  member 
of  the  equation  shall  be  the  original  binomial  (1).     Thus, 

.   x'^  —  4tax  z=i  {x  —  2ay  —  4^^. 
The  original  binomial  is  now  expressed  as  the  difference  of 
two  squares.     Therefore,  the  above  process  is  a  solution  of  the 
problem :    Having  a  binomial  of  which  one  term  is  a  ^perfect 
square,  to  express  it  as  a  difference  of  two  squares, 

EXERCISES. 

Express  the  following  binomials  as  differences   of    two 
squares : 


I 

3 
5 
7 
9 

IT 

13 


x'^  +  2xy,  2.  x^  -\-  4:xy, 

x^  +  ^ax.  4.  4a;2  -f  ^xy, 

4:X^  +  4:xy.  6.  dx^  -\-  ax, 

IGx^  +  S2mx,  8.  x^  +  4.x, 

a\r?  +  2a^x,  10.  Wx^  +  2. 

m%2  +  1.  12.  ^p^x^  +  Ix, 

4:X^  '  9a^x^ 


IRRATIONAL   FACTORS.  169 

Irrational  Factors. 

188.  When  we  introduce  surds,  many  expressions  can  be 
factored  which  have  no  rational  factors.  The  following 
theorem  may  be  applied  for  this  purpose : 

Theorem.  The  difference  of  any  two  quantities  is 
equal  to  the  product  of  sum  and  difference  of  thek 
square  roots. 

In  the  language  of  algebra,  if  a  and  h  be  the  quantities,  we 
shall  have 

which  can  be  proved  by  multiplying  and  by  §  80,  (3). 

EXERCISES. 


Factor 

I.     m  —  n. 

2. 

m  —  1. 

3.     am  —  hn. 

4- 

4«2m  _  9, 

5.     x^  —  m. 

6. 

x^  —  {m  -f  n\ 

7.     {x  —  af  - 

1  , 
--{m- 

-n). 

8. 

x^  —  (in  —  n). 

9.     {a  +  hf  —  (4;/  __  qy     jq^     -^2 _j_  ^xy  +  y^  —  {m  -f  w)i 

Find  the  irrational  square  roots  of  the  following  expressions 
by  the  principles  of  §  186  : 

11.  a  —  2  -f  a~K  Ans,  a^  —  a~^. 

12.  X  —  2Vxy  +  y,  13.     4  -{-  4a/3  -f  3. 
14.     9  +  5  —  6a/5.  15.     4a  4-  ^  —  4tah^. 
16.     fl  +  ^»  +  2(a4-Z^)ir(;-f  2:1    17.     3  +  2a/15  +  5. 

18.     3  +  5  —  2a/15. 

20.     a  —  2\/a  +  1.        ^ 

i        1 

22.     a  +  2^3  4-    1. 

a      a      a 

24.  4  +  3  +  9- 

26.     ^5  _|.  2  -f  a -5. 
28.     a  -\-  b  —  4:  -\- 


19. 

X  y  Vxy 
4  "^4         2    ■ 

21. 

«  —  2a^  4-  ai 

23. 

7           .    ai 

25- 

-  +  -  +  -. 
16  ^  4  ^  4 

27. 

4^:3  _  8  -t-  4a;-8 

a  -\-h 


BOOK    VI. 

EQUATIONS   REQUIRING  IRRA 
TIONAL     OPERATIONS. 


CHAPTER    I. 
EQUATIONS    WITH    TWO    TERMS    ONLY. 

189.  In  the  present  chapter  we  consider  equations  which 
contain  only  a  single  power  or  root  of  the  unknown  quantity. 

Such  an  equation,  when  reduced  to  the  normal  form,  will 
be  of  the  form 

Ax^  +  ^  =  0. 

By  transposing  B,  dividing  by  A,  and  putting 

B 

the  equation  may  be  written, 

x^  —  a  =  0. 
or  x^  =i  a,  (1) 

Here  n  maybe  an  integer,  or  it  may  represent  some  fraction. 
Such  an  equation  is  called  a  Binomial  Equation,  because 
the  expression  x'^  —  a  is  a  binomial. 

Solution  of  a  Binomial  Equation. 

190.  1.  When  the  exponent  ofx  is  a  ivhole  numher.  If  we 
extract  the  n^^  root  of  both  members  of  the  equation  (1),  these 
roots  will,  by  Axiom  Y,  still  be  equal.  The  n^^  ri8f6t  of  x'^  being 
X,  and  that  of  a  being  a^,  we  have  ^ 

X  =  a% 
and  the  equation  is  solved. 


BINOMIAL    EQUATIONS,  171 

2.  When  the  exjjonent  is  fractional.    Let  the  equation  be 

m 

^  —  a. 
Raising  both  members  to  the  n^^  power,  we  have 

x^  =  aP', 
Extracting  the  m^^  root, 

X  —  a^. 

If  the  numerator  of  the  exponent  is  unity,  we  only  have  to 
suppose  m  =  1,  which  will  give 

X  =  a^. 

Hence  the  binomial  equation  always  admits  of  solution  by 
forming  powers,  extracting  roots,  or  both. 

Special  Forms  of  Binomial  Equations. 

Bef,  When  the  exponent  n  is  an  integer,  the  equa- 
tion is  called  a  Pure  Equation  of  the  degree  n. 

When  71  =  2^  the  equation  is  a  Pure  Quadratic 
Equation. 

When  n  =  3,  the  equation  is  a  Pure  Cubic  Equa- 
tion. 


EXERCISES. 

Find  the  values  of  x  in  the 

following  equations : 

I. 

P 

x^ 

Ans.    X  =  -' 

2. 

a  -^  h 

^     =c. 
■  x^ 

3. 

a                h 

xi  —  b       xi  —  a 

4. 

9         x^ 
a;  ""  24' 

5- 

x  —  2a       2x—  b 

X  —  a   ~~  x—b 

6 

x^  —  na       nx^  —  h 

7- 
9- 

a  At  0           xa. 

x^  —  a   ~    x^  —  l 
yi        xi 

^  "«-* 

8. 

\^x  -\-  a^          b  —  a 
a  +  b          ^x  —  a^ 

172  POSITIVE  AND   NEGATIVE  BOOTS. 

In  the  last  example,  clearing  the  equation  of  fractions,  we  shall  have 


or  {x^  -  a'^)^  =  62  _  a\ 

We  square  both  sides  of  this  equation,  which  gives  another  in  which 
a*^  only  appears. 

lo.     {x  —  a)i  =  h^.  II.     {x^  —  a^p  =  mx. 

12.     (V^  —  'v/^)i  =  nx^. 

Positive  and  Neg^ative  Roots. 

191.  Since  the  square  root  of  a  quantity  may  be  either 
positive  or  negative,  it  follows  that  when  we  have  an  equation 
such  as 

x^  z=z  a, 

and  extract  the  square  root,  we  may  have  either 

X  =  +  a^, 
or  X  =z  —  a^. 

Hence  there  are  two  roots  to  every  such  equation,  the  one 
positive  and  the  other  negative.  We  express  this  pair  of  roots 
by  writing 

X  z=z  ±a^, 

the  expression  ±  ^^  meanr.ig  either  +  a^  or  —  a^. 

It  might  seem  that  since  the  square  root  of  x^  is  either  +ir  or  —x,  we 
should  write 

having  the  four  equations, 


±  X 

- 

±aK 

X 

z= 

ai 

X 

= 

— 

aK 

—  X 

= 

+ 

ai, 

—  X 

= 

— 

aK 

But  the  first  and  fourth  of  these  equations  give  identical  values  of  x 
by  simply  changing  the  sign,  and  so  do  the  second  and  third. 

PROBLEMS     LEADINO    TO    PURE    EQUATIONS. 

1.  Find  three  numbers,  such  that  the  second  shall  be 
double  the  first,  the  third  one-third  the  second,  and  the  sum 
of  their  squares  196. 

2.  The  sum  of  the  squares  of  two  numbers  is  369,  and  the 
difference  of  their  squares  81.     What  are  the  numbers? 


I 


PROBLEMS.  173 

3.  A  lot  of  land  contains  1645  square  feet,  and  its  length 
exceeds  its  breadth  by  12  feet.  What  are  the  length  and 
breadth  ? 

To  solve  this  equation  as  a  binomial,  take  the  mean  of  the  length 
and  breadth  as  the  unknown  quantity,  so  that  the  length  shall  be  as  much 
greater  than  the  mean  as  the  breadth  is  less. 

4.  Find  a  number  such  that  if  9  be  added  to  and  subtracted 
from  it,  the  product  of  the  sum  and  difference  shall  be  175. 

5.  Find  a  number  such  that  if  a  be  added  to  it  and  sub- 
tracted from  it  the  product  of  the  sum  and  difference  shall  be 
2a  +  1. 

6.  One  number  is  double  another,  and  the  difference  of 
their  squares  is  192.     What  are  the  numbers  ? 

7.  One  number  is  8  times  another,  and  the  sum  of  their 
cube  roots  is  12.     What  are  the  numbers  ? 

8.  Find  two  numbers  of  which  one  is  3  times  the  other, 
and  the  square  root  of  their  sum,  multiplied  by  the  lesser,  is 
equal  to  128. 

9.  What  two  numbers  are  to  each  other  as  2:3,  and  the 
sum  of  their  squares  =  325  .^ 

Note.     If  we  represent  one  of  the  numbers  by  2x,  the  other  will  be  ^x. 

10.  What  two  numbers  are  to  each  other  as  m\ny  and 
the  square  of  their  difference  equal  to  their  sum  ? 

11.  What  two  numbers  are  to  each  other  as  9  to  7,  and  the 
cube  root  of  their  difference  multiplied  by  the  square  root  of 
their  sum  equal  to  16  ? 

12.  Find  X  and  y  from  the  equations 

ax^  -f  hy'^  =  c, 
a!x^  4-  Vy'^  =  c'. 

13.  The  hypothenuse  of  a  right-angled  triangle  is  26  feet 
in  length,  and  the  sum  of  the  sides  is  34  feet.     Find  each  side. 

Note.  It  is  shown  in  Geometry  that  the  square  of  the  hypothenuse 
of  a  right-angled  triangle  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sides.  In  the  present  problem,  take  for  the  unknown  quantity  the 
amount  by  which  each  unknown  side  differs  from  half  their  sum. 

14.,  Two  points  start  out  together  from  the  vertex  of  a 
right  angle  along  its  respective  sides,  the  one  moving  m  feet 
per  second  and  the  other  7i  feet  per  second.  How  long  will 
they  require  to  be  c  feet  apart  ? 

15.  By  the  law  of  falling  bodies,  the  distance  fallen  is  pro- 
portional to  the  square  of  the  time,  and  a  body  falls  16  feet 
the  first  second.     How  long  will  it  require  to  fall  h  feet  ? 


174  QUADBATIG  EQUATIONS. 

CHAPTER     II. 

QUADRATIC     EQUATIONS. 

193.  Def.  A  Quadratic  Equation  is  one  which, 
when  reduced  to  the  normal  form,  contains  the  second 
and  no  higher  power  of  the  unknown  quantity. 

A  quadratic  equation  is  the  same  as  an  equation  of  the  second  degree. 

Def.  A  Pure  quadratic  equation  is  one  which  con- 
tains the  second  power  only  of  the  unknown  quantity. 

The  treatment  of  a  pure  quadratic  equation  is  given  in  the  preceding 
chapter. 

Def.  A  Complete  quadratic  equation  is  one  which 
contains  both  the  first  and  second  powers  of  the  un- 
known quantity. 

The  normal  form  of  a  complete  quadratic  equation  is 

ax^  -^  bx  -\-  c  =  0.  (1) 

If  we  divide  this  equation  by  a,  we  obtain 

(2) 


x^  +  ^-x 
a 

a 

Putting, 

for  brevity,      -  = 

--P^ 

c  _ 
a  ~ 

-<1, 

equation 

will  be  written  in 

the  form, 

x^  -\-px  +  q  =1  0. 

(3) 

Def.    The  equation 

x^  -^  px  ■\-  q  -^^ 
is  called  the  General  Equation  of  the  Second  Degree, 
or  the  General  Quadratic  Equation,  because  it  is  the 
form  to  which  all  such  equations  can  be  reduced. 


QUADRATIC   EQUAT10N8,  175 


Solution  of  a  Complete  Quadratic  Equation. 

193.  A  quadratic  equation  is  solved  hy  adding  such 
a  quantity  to  its  two  members  that  the  inember  contain- 
ing the  unhnoiun  quantity  shall  he  a  perfect  square, 
(§187.) 

We  first  transpose  q  in  the  general  equation,  obtaining 
x^  +  px  =z  —  g. 

We  then  add  j-  to  both  members,  making 

x^  -\-px+^  ='-  —  q. 

The  first  member  of  the  equation  is  now  a  perfect  square. 
Extracting  the  square  roots  of  both  sides,  we  have 


=  Wi 


,+l-^./«= 


From  this  equation  we  obtain  a  value  of  x  which  may  be 
put  in  either  of  the  several  forms, 


X  =z 

-iwr- 

X  = 

p       Vp^  —  4:q 

2^          2 

1 

If  instead  of  substituting  jt)  and  q,  we  treat  the  equation  in 
the  form  (2)  precisely  as  we  have  treated  it  in  the  form  (3),  we 
sliall  obtain  the  several  results, 

_  --b±  V{I^  -^  4:ac) 
"~  2a 


176  QUADRATIC   EQUATIONS. 

194.  The  equation  in  the  normal  form,  (1),  may  also  be 
solved  by  the  following  process,  which  is  sometimes  more  con- 
venient. Transposing  ^/and  multiplying  the  equation  by  a, 
we  obtain  the  result 

a^x^  +  dbx  =  —  ac. 

To  make  the  first  member  a  perfect  square,  we  add  —  to 
each  member,  giving 

a^x^  +  ahx  +  —  n^ ac, 

4        4 

Extracting  the  square  root  of  both  sides,  ive  have 

^^  +  o  =  ^  V(^^  —  ^ac), 
from  which  we  obtain  the  same  value  of  x  as  before. 

195.  Since  the  square  root  in  the  expression  for  x  may  be 
either  positive  or  negative,  there  will  be  two  roots  to  every 
quadratic  equation,  the  one  formed  from  the  positive  and  the 
other  from  the  negative  surds.  If  we  distinguish  these  roots 
with  x^  and  x^^  their  values  will  be 

_  —  ^  —  V{b^  —  iac)   ' 
^^  -  2a 

We  can  always  find  the  roots  of  a  gi  ven  quadratic  equation  by  sub- 
stituting the  coefficients  in  the  preceding  expression  for  x.  But  the  stu- 
dent is  advised  to  solve  each  separate  equation  by  the  process  just  given, 
which  is  embodied  in  the  following  rule : 

I.  Reduce  the  equation  to  its  normal  or  its  general 
fornh,  as  may  he  inmost  convenient, 

II.  Transpose  the  terms  which  do  not  contain  x  to  the 
second  member, 

III.  //  the  coefficient  of  x^  is  unity,  add  one-fourth 
the  square  of  the  coefficient  of  x  to  both  manbers  of  the 
equation  and  exti^act  the  square  root, 

IV.  //  the  coefficient  of  x^  is  not  unity,  either  divide 
by  it  so  as  to  reduce  it  to  unity ^  or  multiply  all  the  terin^ 


QUADBATIG  EQUATIONS.  177 

hy  such  a  factor  that  it  shall  become  a  perfect  square, 
and  complete  the  square  hy  the  rule  0/  §  187. 

EXAMPLE. 
Solve  the  equation 

•      x-l    _^ 

Clearing  of  fractions  and  transposing,  we  find  the  equation  to  become 

2x^--4:lx  +  1  =  0,  (5) 

,       41:?;              1 
^4 — , 

2  -    2 

Adding  J  the  square  of  the  coefficient  of  x  to  each  side,  we  have 
2      11  1681  _  1681       1  _  1673 

^  ■"  2  '"^  "^    16    "^    16         2  ~"    16  ' 
Extracting  the  square  root  and  reducing,  we  find  the  values  of  x  to  he 

X,  =  ^(41  + V1673), 

and 

x^  =  j(41-\/l673). 

Using  the  other  method,  in  order  to  avoid  fractions,  we  multiply  the 
equation  (5)  by  3,  making  the  equation, 

4^2  _  822;  =  ->  2. 

Adding  -j-  ~  —7—  to  each  side  of  the  equation,  we  have 

.  0       o..         4:12        1681  1673 

4a;2  —  82^;  +  -—  ^  — j 2  =  — ; — 

4  4  4 

Extracting  the  square  root, 

41  /1673        ^1673 

2a; 


/1673 


2  "■  r    4  2     ' 

whence  we  find 

41  ±  \/l673 

the  same  result  as  before. 

EXERCISES. 

Reduce  and  solve  the  following  equations  . 

-^  +  2  _  2;  — 2  _  6  ^  +  4      y  —  i  _  10^ 

^'     ,T-.2       :^+~2  "^  6'  ^*     i^--.4  ^  ?/ +  4  *"*   3' 

12 


178  QUADBATIG  EQUATIONS, 


1  2       _  4 

4.     y^  —  %ay  +  ^2  _  ^2  _  q. 
1 1       1       1 


X  +  a 


1  + 


^-'^  =3. 


X  —  a 


X  -{-  a 

y 

2  +  2/      2/'-4'2-^v 

y  -{-  a      y  —  a  _      1  1 


8-     ^rfT-.-;r--  +  ^r— .  =  4. 


«  -f-  iz;       a  —  i?; 


+  3  =  0. 


y  —  a      y  -\-  ci       y  —  ci'      y^  —  a^      V  —  (^ 

X  X 


PROBLEMS. 

1.  Find  two  numbers  such  that  their  difference  shall  be 
6  and  their  product  567. 

2.  The  difference  of  two  numbers  is  6,  and  the  difference 
of  their  cubes  is  936.     What  are  the  numbers  ? 

3.  Divide  the  number  34  into  two  such  parts  that  the 
sum  of  their  squares  shall  be  double  their  product  ? 

4.  The  sum  of  two  numbers  is  60,  and  the  sum  of  their 
squares  1872.     What  are  the  numbers  ? 

5.  Find  three  numbers  such  that  the  second  shall  be  5 
greater  than  the  first,  the  third  double  the  second,  and  the 
sum  of  their  squares  1225. 

6.  Find  four  numbers  such  that  each  shall  be  4  greater 
than  the  one  next  smaller,  and  the  product  of  the  two  lesser 
ones  added  to  the  product  of  the  two  greater  shall  be  312. 

7.  A  shoe  dealer  bought  a  box  of  boots  for  $210.  If  there 
had  been  5  pair  of  boots  less  in  the  box,  tlfey  would  have  cost 
him  %\  per  pair  more,  if  he  had  still  paid  $210  for  the  whole. 
How  many  pair  of  boots  were  in  the  box  ? 

Rem.  If  we  call  x  the  number  of  pairs,  the  price  paid  for  each  pair 
must  have  been 


QUADRA  TIG  EQUATIONS.  179 

8.  A  huckster  bought  a  certain  number  of  chickens  for 
$10,  and  a  number  of  turkeys  for  $15.75.  There  were  4  more 
chickens  than  turkeys,  but  they  each  cost  him  35  cents  a  piece 
less.     How  many  of  each  did  he  buy? 

9.  A  farmer  sold  a  certain  number  of  sheep  for  $240.  If 
he  had  sold  a  number  of  sheep  3  greater  for  the  same  sum,  he 
would  have  received  $4  a  piece  less.  How  many  sheep  did  he 
sell? 

10.  A  party  having  dined  together  at  a  hotel,  found  the 
bill  to  be  $9.60.  Two  of  the  number  having  left  before  pay- 
ing, each  of  tlie  remainder  had  to  pay  24  cents  more  to  make 
up  the  loss.     What  was  the  number  of  the  party? 

11.  A  pedler  bought  $10  worth  of  apples.  30  of  them 
proved  to  be  rotten,  but  he  sold  the  remainder  at  an  advance 
of  2  cents  each,  and  made  a  profit  of  $3.20.  How  many  did 
he  buy  ? 

12.  In  a  certain  number  of  hours  a  man  traveled  48  miles  ; 
if  he  had  traveled  one  mile  more  per  hour,  it  would  have  taken 
him  4  hours  less  to  perform  his  journey ;  how  many  miles  did 
he  travel  per  hour  ? 

13.  The  perimeter  of  a  rectangular  field  is  160  metres,  and 
its  area  is  1575  square  metres.  What  are  its  length  and 
breadth  ? 

14.  The  length  of  a  lot  of  land  exceeds  its  breadth  by 
a  feet,  and  it  contains  m^  square  feet.  What  are  its  dimen- 
sions ? 

15.  A  stage  leaves  town  A  for  town  B,  driving  8  miles  an 
hour.  Three  hours  afterward  a  stage  leaves  B  for  A  at  such  a 
speed  as  to  reach  A  in  18  hours.  They  meet  when  the  second 
has  driven  as  many  hours  as  it  drives  miles  per  hour.  What 
is  the  distance  between  A  and  B  ? 

Note.  The  solution  is  very  simple  when  the  proper  quantity  is  taken 
as  unknown. 


Equations  which   may  be   Reduced   to  Quad- 
ratics. 

196.  Whenever  an  equation  contains  only  two 
powers  of  the  unknown  quantity,  and  the  index  of  one 
power  is  double  that  of  the  other,  the  equation  can  be 
solved  as  a  quadratic. 


180  QTTADBATIG  EQdATIONS. 

Special  Example.    Let  us  take  the  equation 

7^  +  M  +  c  =z  0.  (1) 

Transposing  c  and  adding  -Vi  to  each  side  of  the  equation, 
it  becomes 

4  4 

The  first  member  of  this  equation  is  a  perfect  square, 
namely,  the  square  of  a:^  +  -  J.  Extracting  the  square  roots 
of  both  members,  we  have 

^  +  |S  =  /(jS^  -  c)  =  ±  J  V(52  -  4c). 

Hence,  ^  =  \[-i±  V{V  -  4c)]. 

Extracting  the  cube  root,  we  have 

x=z  -\-h±V(P-4:c)'\^. 

General  Form,  We  now  generalize  this  solution  in  the 
following  way.  Suppose  we  can  reduce  an  equation  to  the 
form 

ax^^  +  hx^  +  c  =  0, 

in  which  the  exponent  n  may  be  any  quantity  whatever,  entire 
or  fractional.     By  dividing  by  a,   transposing,  and  adding     J 

1  7»2 

-  -7,  to  both  sides  of  the  equation,  we  find 
4a^  ^  ' 

a  4«2       4«^      a 

The  first  side  of  this  equation  is  the  square  of 

Hence,  by  extracting  the  square  root,  and  reducing  as  in 
the  general  equation,  we  find 


QUADRATIC   EQUATIONS.  181 

Extracting  the  n^^  root  of  both  sides,  we  have 
X  =  -j— L  [—b±  V(b^  —  4:ac)]K 


""  \  2a  / 


If  the  exponent  ^  is  a  fraction,  the  same  course  may  be 
followed. 

Suppose,  for  example, 

axi  +  ix^  +  c  =  0. 

Dividing  by  a  and  transposing,  we  have 

4     .      J     2  c 

xt  J x'^  = 

a  a 

Adding  —^  to  both  sides, 

4    .    &    2    .      J^  W-         c 

a  Aa^       4:0^      a 

The  left-hand  member  of  this  equation  is  the  square  of 

Extracting  the  square  root  of  both  members, 

I      _^  _  /^  _  c\*  __  {b^  —  ^ac)^ , 


whence,  a;^  =:_ 

Eaising  both  sides  of  this  equation  to  the  f  power,  w^have 

2a 


X  = 


-]^ 


EXERCISES. 


1.  Eind  a  number  which,  added  to  twice  its  square  root, 
will  make  99. 

2.  What  number  will  leave  a  remainder  of  99  when  twice 
its  square  root  is  subtracted  from  it. 


182  QUADBATIG  EQUATIONS. 

3.  One-fifth  of  a  certain  number  exceeds  its  square  root 
by  30.     What  is  the  number  ? 

4.  What  number  added  to  its  square  root  makes  306  ? 

5.  If  from  3  times  a  certain  number  we  subtract  10  times 
its  square  root  and  96  more,  and  divide  the  remainder  by  the 
number,  the  quotient  will  be  2.    What  is  the  number  ? 

Solve  the  equations : 

6.  \y^  —  2^2  _  15.  7.     3^  _  7^2  _  25. 

8.  5«/4  —  ^y^  =  13, 

mm-. 

9.  {x^  +  a^Y"  —  4:{x^  +  a^)^  —  «2  _  2  +  -  • 

€1 

197.  When   the   unknown  quantity  appears  in  the  form 

x^  -\ — 2,  the  square  may  be  completed  by  simply  adding  2  to 

^  .  1     . 

this  expression,   because    x^  -\-  2  -\-  -^    is   a    perfect    square, 

1  ^ 

namely,  the  square  of  x  +  -*    The  value  of  x  may  then  be 

X 

deduced  from  it  by  solving  another  quadratic  equation. 

3 
Example.  Zx^  +  "2  "=  ^^• 

We  first  divide  by  3  and  add  2  to  each  side  of  the  equation, 
obtaining  ,  ,  ^      1        ^^   .   o       ^^ 

^+^  +  ^2  =  y  +  ^  =  y 

Extracting  the  square  root  of  both  sides, 

,   1       2V7       2^/21       2    .., 
X        ^^  3  3 

By  multiplying  by  x,  this  equation  becomes  a  quadratic, 
and  can  be  solved  in  the  usual  way. 

Let  us  now  take  this  equation  in  the  more  general  form, 

1 

x-\--  =  e,  (a) 

2 
which  reduces  to  the  foregoing  by  putting  e  = -^  ^21.     Clear* 

ing  of  fractions  and  transposing, 

x^  —  ex  -}-  1  =z  0  ; 


QUADRATIC   EQUATIONS.  183 

which  being  solved  in  the  usual  way,  gives 

e  ±  V(e^  —  4) 
X  = -^ 

The  two  roots  are  therefore 

""^  -  2       ~     ' 

_  e  —  V{e^  —  4) 
^3  -  2 

If  in  the  first  of  these  equations  we  rationalize  the  numer- 
ator  by  multiplying  it  by  e  —  \/{e^  —  4)  (§  185),  we  shall  find 

2  1 

it  to  reduce  to ,  that  is,  to  — •    Therefore, 

e  -  V(e2  -  4)  ^2 

iCj  r=  —  identically. 

Vice  versa,  x^  is  identically  the  same  as  — 
This  must  be  the  case  whenever  we  solve  an  equation  of  the 
form  {a),  that  is,  one  in  which  the  value  oi  x  +  -  is  given. 

X 

50 
Let  us  suppose  first  that  e  =  — ,  so  that  the  equation  is 

1       50 

It  is  evident  that  a;  =  7  is  a  root  of  this  equation,  because 
;    when  we  put  7  for  x,  the  left-hand  member  becomes  7  +  ^r, 

50  1 

which  is  equal  to  —•    If  we  put  ^  for  x,  the  left-hand  mem- 
ber will  become 

7^1         7  ^  ' 


Hence  x  and  -  exchange  values  by  putting  -  instead  of  7, 

X  i 

so  that  their  sum  x  -\-  -  remains  unaltered  by  the  change. 


184  QUADRATIC   EQUATIONS. 

The  general  result  may  be  expressed  thus : 

Because  the  value  of  the  expression  x  +  -  remains  un- 
altered when  we  change  x  into  ,  therefore  the  reciprocal  of 
any  root  of  the  equation 

1 

X  +  -  =  e 

X 

is  also  a  root  of  the  same  equation. 

EXERCISES. 

Find  all  the  roots  of  the  following  equations  without  clear- 
ing the  given  equations  from  denominators : 

1         17  1 

I.     iz;2  _|_       — .  2.     a^x^  +  --^  =  7)1^  —  2. 

x^        4  a^x^ 

3.     16^/2  +  ^3  =  28.  4.     j,-{-y'  =  2m^ 

5.  Show,  without  solving,  that  if  r  be  any  roofc  of  the 
equation  •  1 

^'  +  ^2  =  «> 

then  —  r,  - ,  and will  also  be  roots. 


Factorings  a  Quadratic  Equation. 

198.    1.  Special  Case,     Let  us  consider  the  equation 
a;2_22;— 15  =  0, 
or  a;2  —  2a;  +  1  —  16  =  0, 

or  {x  -  1)2  _  42  =  0. 

Factoring,  it  becomes  (§  90), 

(ic  —  1  +  4)  {x  —  1  —  4)  =  0, 
or  {x  +  3)  {x  —  5)  =:  0. 

Therefore  the  original  equation  can  be  transformed  into 
(x  +  3)  (a;  -~  5)  =  0, 

a  result  which  can  be  proved  by  simply  performing  the  multi- 
plications. 


qUJiDBATIG  EQUATIONS,  185 

This  last  equation  may  be  satisfied  by  putting  either  of  its 
factors  equal  to  zero  ;  that  is,  by  supposing 

a;  +  3  =  0,     whence    a;  =  —  3  ; 

or  ;c  —  5  =  0,     whence    :r  =  +  5. 

These  are  the  same  roots  which  we  should  obtain  by  solving 
the  original  equation. 

2.  Factoring  the  General  Quadratic  Equation,  Let  us  con- 
sider the  general  quadratic  equation, 

x^  -\-  px  -\-  q  z=  0.  {a) 

!N'ow,  instead  of  thinking  of  a;  as  a  root  of  this  equation, 
let  us  suppose  x  to  have  any  value  whatever,  and  let  us  con- 
sider the  expression 

x^  +  2^x  -f  qy  (1) 

which  for  shortness  we  shall  call  X,    Let  tis  also  inquire  how 
it  can  be  transformed  without  changing  its  value. 

First  we  add  and  subtract  -^p^,  so  as  to  make  part  of  it  a 
perfect  square.     It  thus  becomes, 

X  =  x^  4-  2^x  +  -;;2  _  _^2  +  ^  . . 
or,  which  is  the  same  tiling, 

Factoring  this  expression  as  in  §  188,  it  becomes 


X  = 


^  +  2P  + 


The  student  should  now  prove  that  this  expression  is  really  equal  to 
x'^  +  px  +  q,  by  performing  the  multiplication. 


Let  us  next  put,  for  brevity, 

1 

"2 


-Ip-i^f-gf. 


186  QUADRATIC   EQUATIONS. 

The  preceding  value  of  X  will  then  become, 

X=  {x-a){x-.  (3),  (3) 

an  expression  identically  equal  to  (1),  when  we  put  for  a  and 
P  their  values  in  (2). 

Let  us  return  to  the  supposition  that  this  expression  is  to 
be  equal  to  zero,  and  that  x  is  a  root  of  the  equation. 

The  equation  (a)  will  then  be 

{x  -a){x-(3)  z=  0.  (4) 

But  no  product  can  be  equal  to  zero  unless  one  of  the  fac- 
tors is  zero.     Hence  we  must  have  either 

X  —  cc  =  0,     whence    x  =  cc-, 
or  X  —  P  =  0,     whence    x  =  fi 

Hence,  a  and  ^  are  the  two  roots  of  the  equation  (a). 

The  above  is  another  way  of  solving  tlio  quadratic 
equation. 

To  compare  the  expressions  (1)  and  (3),  let  us  perform  the 
multiplication  in  the  latter.     It  will  become, 

X  =  x^—  {a  +  P)x  +  ap.        . 

Since  this  expression  is  identically  the  same  as  x^-^px-^q, 
the  coefficients  of  the  like  powers  of  x  must  be  the  same. 
That  is, 

cc  +  P  =  -p,)  .   •  ,  . 

which  can  be  readily  proved  by  adding  and  multiplying  the 
equations  (2). 

This  result  may  be  expressed  as  follows  : 

Theorem,  When  a  quadratic  equation  is  reduced 
to  the  general  form 

x^  ■\-  px  ■\-  q  —  ^^ 
the  coefficient  of  x  will  be  equal  to  the  sum  of  the  roots 
with  the  sign  changed. 

The  term  independent  of  x  will  be  equal  to  the 
product  of  the  roots. 

The  student  may  ask  why  can  we  not  determine  the  roots  of  the 
quadratic  equation  from  equations  (5),  regarding  a  and  /3  as  the  unknown 
quantities  ? 


QUADBATIG  EQUATIONS. 


187 


We  can  do  so,  but  let  us  see  what  tlie  result  will  be.     We  eliminate 
either  «  or  /3  by  substitution  or  by  comparison. 

From  the  second  equation  (5)  we  have, 

a  =  ^-- 

a 

Substituting  this  in  the  first  equation,  we  have 


a  + 


P- 


Clearing  of  fractions  and  transposing, 

«2  +  ^e«  +  ^  =  0. 

We  have  now  the  same  equation  with  which  we  started,  only  a  takes 
the  place  of  x.  If  we  had  eliminated  cc,  we  should  have  had  the  same 
equation  in  /3,  namely, 

/32+^/3  +  g^  =  0. 

So  the  equations  (5),  when  we  try  to  solve  them,  only  lead  us  to  the 
original  equation. 

199.  To  form  a  Quadratic  Equation  when  the  Roots  are 
given.  The  foregoing  principles  will  enable  us  to  form  a  quad- 
ratic equation  which  shall  have  any  given  roots.  We  have 
only  to  substitute  the  roots  for  a  and  /3  in  equation  (4),  and 
perform  the  multiplications. 

EXERCISES. 

Form  equations  of  which  the  roots  shall  be : 


I,  +1  and  —  1. 

3.  —  3  and  —  2. 

5.  7  +  2a/3  and  7— 2^3. 

7.  ~  1  and  +  2. 

9.  +1  and  —  2. 


II. 


-  and  -• 
4  5 


13.     2  +  ^/2  and  2  —  ^/2.      14. 
15-     5  +  7V5  and  5  —  7^/5.    16 


3  and  2. 

3  +  2Vl0and  3-2^10. 

+  1  and  +  2. 

—  1  and  —  2. 

2  +  a/S  and  2  —  ^/b, 

7,9 

2  "^^  2- 

9  +  2\/2  and  9  —  2a/2 

a  -\'h  and  a  —  h. 


17.     a  +  A/a^  —  h^  and  a  —  ^/a^  —  l\ 


188  IMA OINAR  Y  MOOTS. 

Equations  having  Imaginary  Roots. 

300.  When  we  complete  the  square  in  order  to  solve  a 
quadratic  equation,  the  quantity  on  the  right-hand  side  of  the 
equation  to  which  that  square  is  equal  must  be  positive^  else 
there  can  be  no  real  root.  For  if  we  square  either  a  positive 
or  negative  quantity,  the  result  will  be  positive.  Hence,  if 
the  square  of  the  first  member  comes  out  equal  to  a  negative 
quantity,  there  is  no  answer,  either  positive  or  negative,  which 
will  fulfil  the  conditions.  Such  a  result  shows  that  impossible 
conditions  have  been  introduced  into  the  problem. 

E  X  AMPLES. 

1.  To  divide  the  number  10  into  two  such  parts  that  their 
product  shall  be  34. 

If  we  proceed  with  this  equation  in  the  usual  way,  we  shall 
have,  on  completing  the  square, 

x^  —  lOiT  +  25  =  —  9, 
or  {x  —  5)2  =  —  9. 

The  square  being  negative,  there  is  no  answer.  On  con- 
sidering the  question,  we  shall  see  that  the  greatest  possible 
product  which  the  two  parts  of  10  can  have  is  when  they  are 
each  5.  It  is  therefore  impossible  to  divide  the  number  10 
into  two  parts  of  which  the  product  shall  be  more  than  25  ;  and 
because  the  question  supposes  the  product  to  be  34,  it  is  im- 
'possible  in  ordinary  numbers. 

2.  Suppose  a  person  to  travel  on  the  surface  of  the  earth  to 
any  distance ;  how  far  must  he  go  in  order  that  the  straight 
line  through  the  round  earth  from  the  point  whence  he  started 
to  the  point  at  which  he  arrives  shall  be  8000  miles? 

It  is  evident  that  the  greatest  possible  length  of  this  line  is 
a.  diameter  of  the  earth,  namely,  7,912  miles.  Hence  he  can 
never  get  8,000  miles  away,  and  the  answer  is  impossible. 

In  such  cases  the  square  root  of  the  negative  quantity  is 
considered  to  be  part  of  a  root  of  the  equation,  and  because  it 
is  not  equal  to  any  positive  or  negative  algebraic  quantity,  it 
is  called  an  imaginary  root.  The  theory  of  such  roots  will  be 
explained  in  a  subsequent  book. 


I 


IRBATIONAL   EQUATIONS.  189 


CHAPTER    III. 

REDUCTION    OF    IRRATIONAL    EQUATIONS    TO    THE 
NORMAL    FORM. 


201.  An  Irrational  Equation  is  one  in  which  tlio 
unknown  qnantity  appears  under  the  radical  sign. 

An  irrational  equation  may  be  cleared  of  fractions 
in  the  same  way  as  if  it  were  rational. 

Example.     Clear  from  fractions  the  equation 

Vx  +  6?  -f  Vx  —  a  _        2a 
Vx  -\-  a  —  Vx  —  a        Vx^  —  a^ 


Multiplying  both  members  by  Vx^  —  a^  =  \/x  +  aVx—a, 
we  have 

{x  4-  a)  Vx  —  a  -\-  {x  —  a)  Vx  -^  (^  __  ^ 
Vx  -i-  a  —  Vx  —  a 

Next,  multiplying  by  Vx  -{-  a  ^  Vx  —  a,  we  have 


{x-\-a)  Vx  —  a  -\-  {x  —  a)  Vx  -\-  a  =:  2aVx-\-a  —  2aVx  —  a. 
Transposing  and  reducing,  we  have 

{x  +  da)  Vx  —  a  -i-  {x  —  3a)  Vx  +  «  =  0, 
and  the  equation  is  cleared  of  denominators. 

Clearing  of  Surds. 

303.  In  order  that  an  irrational  equation  may  be  solved, 
it  must  also  be  cleared  of  surds  which  contain  the  unknown 
quantity.  In  showing  how  this  is  done,  we  shall  suppose  the 
equation  to  be  cleared  of  denominators,  and  to  be  composed  of 
terms  some  or  all  of  which  are  multiplied  by  the  square  roots 
of  given  functions  of  x. 

Let  us  take,  as  a  first  example,  the  equation  just  found. 
Since  a  surd  may  be  either  positive  or  negative,  the  equation 
in  question  may  mean  any  one  of  the  following  four: 


190  IRRATIONAL    EQUATIONS. 


{x  -f  3«)  ^/x  —  a+  {x  —  3«)  Vx  -\-  a  =  0, 

(1) 

{x  +  da)  "^x  —  a  —  {x  —  oa)  ^/x  +  «  =  0, 

(3) 

—  (x  -\-  ?id)  ^/x  —  a  -\-  (x  —  ?>a)  \/x  -{-  a  =  0, 

(3) 

—  {x  -\-  3a)  "s/ X  —  a  —  {x  —  3a)  ^/x  +  a  =  0. 

(4) 

But  the  third  equation  is  merely  the  negative  of  the  second, 
and  the  fourth  the  negative  of  the  first,  so  that  only  two  have 
different  roots.     Let  us  put,  for  brevity, 


P  —  (x  \3a)^/x  —  a-\-  {x  —  3a)  \/x~^\^,  ) 
Q  =  (x  -\-  3a)  Vx  ^  a—  {x  —  3a)  "s/x  +  r?,  ) 
and  let  us  consider  the  equation, 

PQ  =  0.  (6) 

Since .  this  equation  is  satisfied  when,  and  only  when,  we 

have  either  P  =  0  or  §  =:  0,  it  follows  that  every  value  of  x 

.which  satisfies  either  of  the  equations  (1)  or  (2)  will  satisfy  (G). 

Also,  every  root  of  (6)  must  be  a  root  either  of  (1)  or  (2). 

If  we  substitute  in  (6)  the  values  of  F  and  Q  in  (5),  we 
shall  then  have 

{x  4-  3aY{x  —  a)  —  {x  —  3a)^ {x -{-  a)  =  0, 

which  reduces  to  6x^  —  9a^  =  0, 

,    .  3a 

and  gives  x  =:  ±  — —  • 

V5 

It  will  be  remarked  that  the  process  by  which  we  free  the 
equation  from  surds  is  similar  to  that  for  rationalizing  the 
terms  of  a  fraction  employed  in  §  185. 

As  a  second  example,  let  us  take  the  equation, 


V^  -h  11  +  Vx  —  4:  —  5  =  0.  (a) 

We  write  the  three  additional  equations  formed  by  combin- 
ing the  positive  and  negative  values  of  the  surds  in  every  way: 


•—  a/S~+Ti  +  Vx  —  4:  —  5  r=  0, 
Vx  +  11  —  a/S'^^^  —  5  =  0, 
—  Vx  +  11  —  Vx  —  4:  —  5  =  0. 
The  product  of  the  first  two  equations  is 


IRRATIONAL    EQUATION'S.  191 

(^^/^Z^l^  5)2  _  {x  +  11)  =  0, 
or  10  —  lOVi"^^  =  0.  (1) 

The  product  of  the  last  two  is 

10  +  l0\/^^"4  z=  0.  (2) 

The  product  of  these  two  products  is 

100  —  100  (^  —  4)  =  0, 
which  gives  x  =i  b. 

It  will  be  remarked  that  (2)  differs  from  (1)  only  in  having 
the  sign  of  the  surd  different.  This  must  be  the  case,  because 
the  second  pair  of  equations  formed  from  [a)  differ  from  the 
first  pair  only  in  having  the  sign  of  the  surd  '^/x  —  4  different. 
Hence  it  is  not  necessary  to  write  more  than  one  pair  of  the 
equations  at  each  step.     The  general  process  is  as  follows : 

I.  Change  the  sign  of  one  of  the  surds  in  the  given 
equation,  and  multiply  the  equation  thus  formed  hy  the 
original  equation. 

II.  Reduce  this  product,  in  it  change  the  sign  of  an- 
other of  the  surds,  and  form  a  neiu  product  of  the  two 
equations  thus  formed, 

III.  Continue  the  process  until  an  equation  without 
surds  is  reached. 

Example.     Solve 

A/8a;  +  9  +  V^x -{-  6  +  Va?  +  4  =  0, 


Changing  the  sign  of  ^/x  +  4, 


^/Sx  +  9  +  ^/^x  +  6  —  v'o;  +  4  =  0. 
The  product  is 

(VSx  +  9  +  \/%x  +  6/  —  {x  +  4)  =  0, 
or,  after  reduction, 


9:^  +  11  -!_  2^y^x  +  9  \/'lx  +  6  =  0. 


Changing  the  sign  of  ^/'ilx  -f-  6,  we  have 

^x  4-  11  —  2^/8^  +  9  V^x  +  G  —  0. 


192  IRRATIONAL   EQUATIONS. 

The  product  of  the  last  two  equations  reduces  to 
17^2  _  QQx  —  95  =  0, 

which  being  solved  gives  x  =  — ~ 

Remark.  Equations  containing  surds  may  often  reduce  to  the  form 
treated  in  §  196.  In  this  case,  the  methods  of  that  section  may  be  fol- 
lowed. 

EXERCISES. 

Solve  the  equations : 

1 1 2Va  —  Wx 

'\/x  +  ^a      Vx  —  ^/a  X  —  a 

Vx''-\-a       X 


"s/a^  —X       ^ 

4.  Vx  +  i-i  +  a/uj^^iT=  14. 

5.  (3  —  x)i  —  (3  +  x'^)i  =  0. 

6.  Vet  +  ^x  +  Vet  —  ^x  ■=.  2Vx  +  — • 

Vx-\-2       ^  -  4:       Vx-2 
V'Ex  +  3  2        ' 


Vx  -\-  3  —  Vx--4:  =:1, 


9.     Va^  —  2X  + 


Va^  —  2x 

X  +  Vx       X  (x  —  1) 

10.     —  =  . 

X  —  Vx  ^ 

Vr+a  1 

31.  ; =       , 

Vx  —  a  +  yctx  —  1       Vx  —  1 


SIMVLTANE0U8   QUADEATIG  EQUATIONS.         193 

CHAPTER    IV. 

SIMULTANEOUS    QUADRATIC    EQUATIONS. 

Between  a  pair  of  simultaneous  general  quadratic  equations 
one  of  the  unknown  quantities  can  always  be  eliminated.  The 
resulting  equation,  when  reduced,  will  be  of  the  fourth  degree 
with  respect  to  the  other  -unknown  quantity,  and  cannot  be 
solved  like  a  quadratic  equation. 

But  there  are  several  cases  in  which  a  solution  of  two  equa- 
tions, one  of  which  is  of  the  second  or  some  higher  degree, 
may  be  effected,  owing  to  some  of  the  terms  being  wanting  in 
one  or  both  equations. 

303.  Case  I.  When  one  of  the  equations  is  of 
the  first  degree  only. 

This  case  may  be  solved  thus : 

KuLE.  Find  the  value  of  one  of  the  unhnown  quan- 
tities in  terms  of  the  other  from  the  cquaMon  of  the  first 
decree.  This  value  being  substituted  in  the  other  equa- 
tion, we  shall  have  a  quadratic  equation  from  which  the 
other  unhnown  quantity  may  he  found. 

Example.    Solve 

2iz:2  +  dxy  ^  6y^  —  x -^  6y  i=  26,  ) 

2x^^  =z    5.  )  ^^^ 

From  the  second  equation  we  find 

Whence,  .^  =  V  +  g^+jg. 

Substituting  this  value  in  the  first  equation  and  reducing, 
we  find 

4?/2  +  16y  +  10  =r  26. 

Solving  this  quadratic  equation, 
13 


194         SIMULTANEOUS   QUADBATIG   EQUATIONS 

^=:— 2±V'8=— 2±  2\/2. 
This  value  of  y  being  substituted  in  the  equation  {h)  gives, 
1  -{-  sv^s        —  1  ±  6a/2 


X  =z 


2  ~  2 


The  same  problem  may  be  solved  in  tbe  reverse  order  by  eliminating 
4/  instead  of  x.    Tlie  second  equation  (a)  gives 

2x  —  5 

If  we  substitute  this  value  oiy  in  the  first  equation,  we  shall  have  a 
quadratic  equation  in  x,  from  which  the  value  of  the  latter  quantity  can 
DC  found. 


Solve 

EXERCISES. 

I. 

x^  —  2xy  +  4?/2  =  21. 
2x  -\-  y  =  12. 

2. 

3^2  _  2^2  _|_  5^  _  2^  =  28. 
2;  +  2/  +  4  =  0. 

3- 

X  -{-2y  =  0, 

4. 

3x^  +  2^^  =  813, 
7a;  —  4^  =     17. 

5. 

^  +  y  =  7, 

a:       y  _    '^ 
y~~x~  12' 

304.  Case  II.  When  each  equation  contains 
only  one  term  of  the  second  deffvee^  and  that  term 
''his  the  sayne  2>^oduct  or  square  of  the  unknown 
quantities  in  the  two  equations. 

Such  equations  are 

ax^  -{•  dx  -{-  ey  -{-  f  =  0,  |  ,  . 

a'x^  +  d'x  +  e'y  +/  =  0,  f  ^^^ 

where  the  only  term  of  the  second  degree  is  that  in  a^. 

If  we  eliminate  x'^  from  these  equations  by  multiplying  the 
first  by  a'  and  the  second  by  a,  and  subtracting,  we  have 


SIMULTANEOUS   QUADRATIC   EQUATIONS,         195 

{a'd  —  ad')  x  +  {a!e  —  ae')  y  -\-  a'f  —  af  =  0. 
Bui  iving  this  equation  with  respect  to  x,  we  find 
_  {ae'^a'e)y^  af  -  a'f 
^~  d'd-ad'  ^^> 

By  substituting  this  value  of  x  in  either  of  the  equations 
(a),  we  shall  have  a  quadratic  equation  in  y.  Solving  the 
latter,  we  shall  obtain  two  values  of  y.  Substituting  these  in 
(b),  we  shall  have  the  two  corresponding  values  of  x,  and  tlie 
solution  will  be  complete.     Hence  the  rule, 

Eliminate  the  term  of  the  second  decree  by  addition 
or  subtraction,  and  use  the  resulting  equation  of  the  first 
degree  with  either  of  the  original  equations,  as  in  Case  I, 

Example.     Solve 

2xy  —  4:x-^6y-jzz  23,  ) 

3xy  +  7x  +    y  =  41,)  ^^^ 

Multiplying  the  first  equation  by  3  and  the  second  by  2, 
and  subtracting,  we  have 

—  26x  +  13y  =  —  13  ;  (b) 

whence,  x  =  -y  -\-  --  (c) 

Substituting  this  value  in  tl>e  first  equation,  we  find  a 
quadratic  equation,  which,  being  solved,  gives 

y  =  -2±  V29. 

Substituting  these  values  in  (c),  the  result  is 

The  two  sets  of  values  of  the  unknown  quantities  are 
therefore 

''i^-l  +  l^^^'  ^,  =  -l-lV29. 

y^   =   -2  +  V2d,  y^=:   -2-  a/29. 

We  might  have  obtained  the  same  result  by  solving  the  equation  (c) 
with  respect  to  y,  and  substituting  in  (a).  The  student  should  practice 
both  methods. 


196         SIMULTANEOUS   QUADRATIC  EQUATIONS. 

EXKRCISES. 

1.  %x^  -  dx  —  4y  =  25,  "'  ^' ' ''''' 

x^  -\-2x  —  3y  =  18. 

2.  2y^  +  y  :=  28, 
^2  _|.  ^r^  '-4:y  =  18. 

dxy  ^2x  -i-  6y  =  70. 

305,  Case  III.  When  neitliei^  equation  €on° 
tains  a  term  of  the  first  degree  in  oc  or  y. 

EuLE.  Eliminate  the  constant  terms  by  multiplying 
each  equation  by  the  constant  term  of  the  other,  and 
adding  or  subtracting  the  two  products.  The  result  ivill 
be  a  quadratic  equation,  from  luhich  either  unhnoivn 
quantity  can  be  determined  in  terms  of  the  other.  TJxen 
substitute  as  in  Case  L 


(1) 


Example.     Solve  x^  •{-    xy  —    «/^  =    5, 

2x2  _  3^^  ^  2^2  --  14. 

14  X  1st  eq.,  14a:2  H-  14.xy  —  14?/^  =  70. 

5  X  2d  eq. ,  10:^8  -^  Uxy  -f  10^^  -.  70. 

Subtracting,  ix^  +  29xy  —  24/  =    0. 

This  is  a  quadratic  equation,  by  which  one  unknown  quan- 
tity can  be  expressed  in  terms  of  the  other  without  the  latter 
being  under  the  radical  sign. 

Transposing,  4:X^  +  2dxy  —  24«/l  (2) 

841  1225 

Completing  square,    4,0(^  4-  2^xy  +  'T'^-y^  =  ~T^^^* 

Extracting  root,  2x  -\-  -j-y  :=^  :t-ry* 

wi,                                           —  29  ±  35  3  ^ 

Whence,  x  — ~=- — y  =  jy  or  -—  8y. 

Substituting  the  first  of  these  values  of  x  in  either  of  the 
original  equations,  we  shall  have 

f  =  16 ; 


SIMULTANEOUS   QUADBATIG   EQUATIONS,         197 

whence,  ?/irr±4;        2;=±3. 

SubstiiTuting  the  second  value  of  x^  we  have  • 

II'. 


^2  ^ 


Whence,  ^-±:^;        "*  ^  ^  7lT 

Therefore  the  four  possible  values  of  the  unknown  quanti- 
ties are. 

Each  of  these  four  pairs  of  values  satisfies  the  original 
equation. 

A  slight  change  in  the  mode  of  proceeding  is  to  divide  the 
equation  (2)  by  either  x^  or  y"^,  and  to  find  the  value  of  the 
quotient.     Dividing  by  y^  and  putting 

X 

y 

the  equation  will  become 

4z^2  j^  2du  ^  24  =  0. 
This  quadratic  equation,  being  solved,  gives 

29  i:  35  _  3 

8  *~  4 


IC  =::  —^ —  ^  -   or    —  5, 


X 

Putting  -  for  ic,  and  multiplying  by  y, 

3 

X  =:  -y  ov  —  Sy,  as  before. 

EXERCISES. 

Solve 

1.  x^  —    xy  +    y^  —  3  =  0, 
x^  —  2xy  -f-  4y^  —  4:  =  0. 

2.  2x^  +  3xy  —    1^2  _  2  =  0, 


198         SIMULTANEOUS   QUADRATIG   EQUATIONS. 

306.  Case  IV.  When  the  expressUms  contain- 
inn  tJte  itn known  quantities  in  the  two  equations 
have  comnion  factors. 

Rule.  Divide  one  of  the  equoMons  which  can  he  fac- 
tored by  the  othe7%  and  cancel  the  common  factors. 
Then  clear  of  fractions,  if  necessary,  and  we  shall  have 
an  equation  of  a  lower  degree, 

EXAMPLES. 

1.  x^-\-y^  =  91,     q:  +  y  —  7. 

We  have  seen  (§  94,  Th.  1)  that  x^-\-y^  is  divisible  hy  x-{-y. 
So  dividing  the  first  equation  by  the  second,  we  have 
x^  —  xy  -\-  y'^  =:  13. 
This  is  an  equation  of  the  second  degree  only,  and  when 
combined  with  the  second  of  the  original  equations,  the  solu- 
tion may  be  effected  by  Case  I.     The  result  is, 
:r  =  3  or  4,        y  =  4  or  3. 

2.  xy  +  2/2  =  133,    x^  —  y^  z=z  95. 

Factoring  the  first  member  of  each  equation,  the  equations 
become 

y{x  +  y)  :=.  133,        (x  ^  y)  {x  -  y)  =  95. 

Dividing  one  equation  by  the  other,  and  clearing  of  fractions, 

7 
I2y  =  "ix,     or    y  = —x. 

The  problem  is  now  reduced  to  Case  I,  this  value  of  y 
being  combined  with  either  of  the  original  equations. 

307.  There  are  many  other  devices  by  which  simultaneous 
equations  may  be  solved  or  brought  under  one  of  the  above 
cases,  for  which  no  general  rule  can  be  given,  and  in  which 
the  solution  must  be  left  to  the  ingenuity  of  the  student. 
Sometimes,  also,  an  equation  which  comes  under  one  of  the 
cases  can  be  solved  much  more  expeditiously  than  by  the  rule. 

Let  us  take,  for  instance,  the  equations, 
x%  ^  y2  —  65,        xy  =z  28. 

These  equations  can  be  solved  by  Case  III,  but  the  work 
would  be  long  and  cumbrous.     We  see  that  by  adding  and  ^ 


SIMULTANEOUS   QUADBATIC  EQUATIONS.         199 

subtracting  twice  the  second  equation  to  and  from  the  first, 
we  can  form  two  perfect  squares.  Extracting  the  roots  of 
these  squares,  we  shall  have  two  simple  equations,  which  shall 
give  the  solution  at  once.  Each  unknown  quantity  will  have 
four  values,  namely,  ±  7  ±  4. 

PROBLEMS    AND     EXERCISES. 

The  following  equations  can  all  be  solved  by  some  sliort  and  expe- 
ditious combination  of  the  equations,  or  by  factoring,  without  going 
through  the  complex  process  of  Case  III.  The  student  is  recommended 
not  to  work  upon  the  equations  at  random,  but  to  study  each  pair  until 
he  sees  how  it  can  be  reduced  to  a  simpler  equation  by  addition,  multi- 
plication, or  factoring,  and  then  to  go  through  the  operations  thus  sug- 
gested. 

1.  y^  +  xy  =  14,     x^  +  xy  =  35. 

2.  4:X^  —  2xy  =  208,    2xy  —  y^  =r  39. 

3.  x^  '{-  y  =  4^,     y^  -j-  X  =  4:y. 

If  we  subtract  one  of  these  equations  from  the  other,  the  difference 
will  be  divisible  by  x  —  y. 

4.  .^3  _!_  y^  _|.  3^  _|.  3^  ^  378^    :^  j^y^^Zx  —  Zy  z=z  324. 

5.  x^  +  ^2  ^  74,    x^y  ^  12. 

6.  x^  -^  xy  ■=:  G3,  ,x^  —  y'^  =  77. 

^/x  —  yy 

x^  -\r  xy  =:  a,    y'^  -{-  xy  ^  h, 

x^  +  xy"^  =  10,     y^  +  x^y  =2  5. 

X  =z  aV^  -\-  y,     y  -=^  'b\/x  -f  y, 

x'^x  +  y  r=  12.     y^/x  +  3/  =  15. 

2x^  +  2y^  =  X  -{-  y,     x^  i-  y^  =  x  —  y. 

5x^  —  5y^  =  X  -\-  y,     3x^  —  Sy^  =  x  —  y. 

x^  +  y^-i-z^  =  30,     xy-\-yz-\-zx  =  17,     x  —  y  ^  z  =  2. 


7 

8 
9 

10, 
II 
12 

13 
14 

15 


/x  -\-  y 
x  +  y  —  2\/  — '-^  = 


8 

-y 


200  PROBLEMS. 

1 6.  A  principal  of  $5000  amounts,  with  simple  interest,  to 
$7100  after  a  certain  number  of  years.  Had  the  rate  of  inter- 
est been  1  per  cent,  higher  and  the  time  1  year  longer,  it  would 
have  amounted  to  $7800.     What  was  the  time  and  rate? 

17.  A  courier  left  a  station  riding  at  a  uniform  rate.  Fiye 
hours  afterward,  a  second  followed  him,  riding  3  miles  an 
hour  faster.  Two  hours  after  the  second,  a  third  started  at 
the  rate  of  10  miles  an  hour.  They  all  reach  their  destination 
at  the  same  time.     What  was  its  distance  and  the  rate  of  riding  ? 

18.  In  a  right-angled  triangle  there  is  given  the  hypothe- 
nuse  =  a,  and  the  area  =  Z>^;  find  the  sides. 

19.  Find  two  numbers  such  that  their  product,  sum,  and 
difference  of  squares  shall  be  equal  to  each  other. 

20.  Find  two  numbers  whose  product  is  216;  and  if  the 
greater  be  diminished  by  4,  and  the  less  increased  by  3,  the 
product  of  this  sum  and  difference  may  be  240. 

21.  There  are  two  numbers  whose  sum  is  74,  and  the  sum 
of  their  square  roots  is  12.     What  are  the  numbers  ? 

22.  Find  two  numbers  whose  sum  is  72,  and  the  sum  of 
their  cube  roots  6. 

23.  The  sides  of  a  given  rectangle  are  m  and  n.  Find  the 
sides  of  another  which  shall  have  twice  the  perimeter  and  twice 
the  area  of  the  given  one. 

24.  A  certain  number  of  workmen  require  3  days  to  com- 
plete a  work.  A  number  4  less,  working  3  hours  less  per  day, 
will  do  it  In  6  days.  A  number  6  greater  than  the  original 
number,  working  6  hours  less  per  day,  will  complete  the  work 
in  4  days.  What  was  the  original  number  of  workmen,  and 
how  long  did  they  work  per  day  ? 

25.  Find  two  numbers  whose  sum  is  18  and  the  sum  of 
their  fourth  powers  14096. 

Note.  Since  the  smn  of  the  two  numbers  is  18,  it  is  evident  that 
the  one  must  be  as  much  less  than  9  as  the  other  is  greater.  The  equa- 
tions will  assume  the  simplest  form  when  we  take,  as  the  unknown  quan- 
tity, the  common  amount  by  which  the  numbers  differ  from  9. 

26.  Find  two  numbers,  x  and  y,  such  that 

C(^^y^  :  x^  —  y^    ::    35  :  19, 
xy  =  24. 

27.  Find  two  numbers  whose  sum  is  14  and  the  sum  of 
their  fifth  powers  161294. 


BOOK    VII. 
PROGRESS/OA^S, 

CHAPTER     I. 

ARITHMETICAL      PROGRESSION. 

308.  Def.  When  we  have  a  series  of  numbers  each 
of  which  is  greater  or  less  than  the  preceding  by  a  con- 
stant quantity,  the  series  is  said  to  form  an  Arithmet- 
ical Progression. 

Example.     The  series 

7,  12,  17,  22,  27,  32,  etc.  ; 
7,  5,  3,  1,  —1,  —3,  etc.; 
a  -\-  b,     a,     a  —  h,    a  —  2^,     a  —  3^,    etc., 

are  each  in  arithmetical  progression,  because,  in  the  first,  each 
number  is  greater  than  the  preceding  by  5 ;  in  the  second, 
each  is  less  than  the  preceding  by  2 ;  in  the  third,  each  is  less 
than  the  preceding  by  h. 

Def,  The  amount  by  which  each  term  of  an  arith- 
metical progression  is  greater  than  the  preceding  one  is 
called  the  Common  Difference. 

Def.  The  Arithmetical  Mean  of  two  quantities  is 
half  their  sum. 

AH  the  terms  of  an  arithmetical  progression  except 
the  first  and  last  are  called  so  many  arithmetical  means 
between  the  first  and  last  as  extremes. 

Example.  The  four  numbers,  5,  8,  11, 14,  form  the  four 
arithmetical  means  between  2  and  17. 


202  ABITHMETIGAL    PROGRESSION. 

EXERCISES. 

1.  Form  four  terms  of  the  arithmetical  progression  of 
which  the  first  term  is  7  and  common  difference  3. 

2.  Write  the  first  seven  terms  of  the  progression  of  which 
the  first  term  is  11  and  the  common  difference  —  3. 

3.  Write  five  terms  of  the  progression  of  which  the  first 
term  is  a  —  ^n  and  the  common  difference  27^. 

Problems  in  Progression, 

209.  Let  us  put 

a,  the  first  term  of  a  progression. 
d,  the  common  difference. 
Uy  the  number  of  terms. 
I,  the  last  term. 
2,  the  sum  of  all  the  terms. 
The  series  is  then 

ay    a-\-d,    a-\-2d,  ....?. 

Any  three  of  the  above  five  quantities  being  given,  the 
other  two  may  be  found. 

Pkoblem  I.     Given  the  first  term,  the  cormnon  differ- 
ence, and  the  number  of  terms,  to  find  the  last  term. 
The  1st  term  is  here  a, 

M      "        "  a  +  d, 

3d      "        "  a  +  2d. 

The  coefficient  of  d  is,  in  each  case,  1  less  than  the  number 
of  the  term.  Since  this  coefficient  increases  by  unity  for  every 
term  we  add,  it  must  remain  less  by  unity  than  the  number  of 
the  term.     Hence, 

The  i^^  term  is  a  +  {i  —  1)  d, 
whatever  be  i.     Hence,  when  i  =  n, 

I  =  a  +  {n'^l)d.  (1) 

From  this  equation  we  can  solve  the  further  problems  : 

Problem  II.  Given  the  last  term  I,  the  common  dif- 
ference d,  and  the  number  of  terms  ^i,  to  find  the  first 
term. 


ARITHMETICAL    PR0ORES8I0N.  203 

The  solution  is  found  by  solving  (1)  with  respect  to  a, 

which  gives 

a  =  l-{n^  1)  d.  (2) 

Problem  III.  Given  the  first  and  last  terms,  a  and  I, 
and  the  number  of  terms  n,  to  find  the  common  differ- 
ence. 

Solution  from  (1),  d  being  the  unknown  quantity, 

d  =  ^.  (3) 

71—1  ^    ' 

Problem  IV.  Given  the  first  and  last  terms  and  the 
common  difference,  to  find  the  niimher  of  terms. 

Solution,  also  from  (1), 

I  —  ct  ^   ^        I  —  a  +  d  ... 

«  =  -^-  +  l  =  ^—  (4) 

Problem  V.  To  find  the  sum  of  all  the  terms  of  an 
arithmetical  progression. 

We  have,  by  the  definition  of  S, 

2  =  a  -f  («5  +  cO  -^  («  +  ^^)  + {l  —  d)  +  l, 

the  parentheses  being  used  only  to  distinguish  the  terms. 

Now  let  us  write  the  terms  in  reverse  order.  The  term 
before  the  last  is  ?  —  d,  the  second  one  before  it  I  —  2dy  etc. 

We  therefore  have, 

^  =,1  j^  (l  —  d)  -\-  {l  —  '^d) -{■  {a  -\-  d)  -\-  a. 

Adding  these  two  values  of  2  together,  term  by  term,  we 
find 
22  =  («  +  ?)  +  {a  +  l)  +  {a  +  l)  + +  («  +  0  +  («4-0, 

the  quantity  {a  + 1)  being  written  as  often  as  there  are  terms, 
that  is,  71  times.    Hence, 

22  =  72  {a  +  l\ 

j:  =  7i—-'  (5) 

Eemark.  The  expression  — -— ,  that  is,  half  the  sum  of 
the  extreme  terms,  is  the  mean  value  of  all  the  terms.     The 


204  ARITHMETICAL   PROGRESSION. 

sum  of  the  n  terms  is  therefore  the  same  as  if  each  of  them 
had  this  value. 

310.  In  the  equation  (5)  we  are  supposed  to  know  the 
first  and  last  terms  and  the  number  of  terms.  If  other  quan- 
tities are  taken  as  the  known  ones,  we  have  to  substitute  for 
some  one  of  the  quantities  in  (5)  its  expression  in  one  of  the 
equations  (1),  (2),  (3),  or  (4).  Suppose,  for  example,  that  we 
have  given  only  the  last  term,  the  common  difference,  and  the 
number  of  terms,  that  is,  /,  d,  and  n.  We  must  then  in  (5) 
substitute  for  a  its  value  in  (2).     This  will  give, 

^  =z  nil — -  d\  =z  nl ^—- d.  (6) 

EXERCISES. 

In  arithmetical  progression  there  are 

1.  Given,  common  difference,  +  3;  third  term  =  10. 
Find  first  term.  Ayis,    First  term  —  4. 

2.  Given  4th  term  =  d,  common  difference  =  —  c. 
Find  first  7  terms,  their  sum  and  product. 

3.  Given  3d  term  z=  a  ^  h,  4th  term  =  a  +  2^. 
Find  first  5  terms. 

4.  Given  1st  term  ■=:  a  —  h,  9th  term  =:  9^  +  Ih. 
Find  2d  tprm  and  common  difference. 

5.  Given,  sum  of  9  terms  =  108. 

Find  middle  term  and  sum  of  1st  and  9th  terms. 

6.  Given  5th  term  =r  7^^  —  by,  7th  term  =  9.r  —  ^y. 
Find  first  7  terms  and  common  difference. 

7.  Given  1st  term  =  12,  50th  term  =  551. 
Find  sum  of  all  50  terms. 

8.  To  find  the  sum  of  the  first  100  numbers,  namely, 

1  +  2+3 +99  +  100. 

Here  the  first  term  a  is  1,  the  last  term  I  100,  and  the  number  of 
terms  100.    The  solution  is  by  Problem  V. 

9.  Find  the  sum  of  the  first  n  entire  numbers,  namely, 

1+2  +  3 +  n. 


ARITHMETICAL    PROGRESSION.  205 

10.  Find  the  sum  of  the  first  n  odd  numbers,  namely, 

1  +  3  +  5 +27^  —  1. 

Here  the  number  of  terms  is  n. 

11.  Find  the  sum  of  the  first  n  even  numbers,  namely, 

2  +  4  +  6 +  2/^. 

12.  In  a  school  of  m  scholars,  the  highest  received  134 
merit  marks,  and  each  succeeding  one  6  less  than  the  one  next 
above  him.  How  many  did  the  lowest  scholar  receive  ?  How 
many  did  they  all  receive  ? 

13.  The  first  term  of  a  series  is  m,  the  last  term  2m,  and 
the  common  difiference  d.     What  is  the  number  of  terms? 

14.  The  first  term  is  Ic,  the  last  term  lOh  —  1,  and  the 
number  of  terms  9.     What  is  the  common  difference  ? 

15.  The  middle  term  of  a  progression  is  s,  the  number  of 
terms  5,  and  the  common  difference  —  K  What  are  the  first 
and  last  terms  and  the  sum  of  the  5  terms  ? 

16.  The  sum  of  5  numbers  in  arithmetical  progression  is 
20  and  the  sum  of  their  squares  120.     What  are  the  numbers  ? 

Note.  In  questions  like  tins  it  is  better  to  take  the  middle  term  for 
one  of  the  unknown  quantities.  The  other  unknown  quantity  will  be 
the  common  difference. 

17.  Find  a  number  consisting  of  three  digits  in  arithmeti- 
cal progression,  of  which  the  sum  is  15.  If  the  number  be 
diminished  by  792,  the  digits  will  be  reversed. 

18.  The  continued  product  of  three  numbers  in  arithmet- 
ical progression  is  640,  and  the  third  is  four  times  the  first. 
What  are  the  numbers  ? 

19.  A  traveller  has  a  journey  of  132  miles  to  perform.  He 
goes  27  miles  the  first  day,  24  the  second,  and  so  on,  travelling 
3  miles  less  each  day  than  the  day  before.  In  how  many  days 
will  he  complete  the  journey  ? 

Here  we  have  given  the  first  term  27,  the  common  difference  —3,  and 
the  sum  of  the  terms  132.  To  solve  this,  we  take  equation  (5),  and  sub- 
stitute for  1  its  value  in  (1).     This  makes  (5)  reduced  to 

a  -{-  a  -^  (n  —  1)  ^  n(n  —  1)  d 

^  —  n ^ '—  =  na-\ ^ 

2,  a,  and  d  are  given  by  the  problem,  and  n  is  the  unknown  quan- 
tity. Substituting  the  numerical  value  of  the  unknown  quantities,  the 
equation  becomes 


206  ARITHMETICAL   PROGRESSION 

This  reduced  to  a  quadratic  equation  in  n,  the  solution  of  which  gives 
two  values  of  n.  The  student  should  explain  this  double  answer  by 
continuing  the  progression  to  11  terms,  and  showing  what  the  negative 
terms  indicate. 

20.  Taking  the  same  question  as  the  last,  only  suppose  the 
distance  to  be  140  miles  instead  of  132.  Show  that  the  answer 
will  be  imaginary,  and  explain  this  result. 

21.  A  debtor  owing  $160  arranged  to  pay  25  dollars  the 
first  month,  23  the  second,  and  so  on,  2  dollars  less  each 
month,  until  his  debt  should  be  discharged.  How  many  pay- 
ments must  he  make,  and  what  is  the  explanation  of  the  two 
answers  ? 

22.  A  hogshead  holding  135  gallons  has  3  gallons  poured 
into  it  the  first  day,  6  the  second,  and  so  on,  3  gallons  more 
every  day.     How  long  before  it  will  be  filled  ?  1 

23.  The  continued  product  of  5  consecutive  terms  is  12320 
and  their  sum  40.     What  is  the  progression  ? 

24.  Show  that  the  condition  that  three  numbers,  p,  q,  and 
r,  are  in  arithmetical  progression  may  be  expressed  in  the  form 

q  —  r 

25.  In  a  progression  consisting  of  10  terms,  the  sum  of  the 
1st,  3d,  5th,  7th,  and  9th  terms  is  90,  and  the  sum  of  the  re- 
maining terms  is  110.     What  is  the  progression  ? 

26.  In  a  progression  of  an  odd  number  of  terms  there  is 
given  the  sum  of  the  odd  terms  (the  first,  third,  fifth,  etc.)^ 
and  the  sum  of  the  even  terms  (the  second,  fourth,  etc.). 
Show  that  we  can  find  the  middle  term  and  the  number  of 
terms,  but  not  the  common  difference. 

27.  In  a  progression  of  an  even  number  of  terms  is  given 
the  sum  of  the  even  terms  =  119,  the  sum  of  the  odd  terms  = 
105,  and  the  excess  of  the  last  term  over  the  first  =  26.  What 
is  the  progression  ? 

28.  Given  a  and  I,  the  first  and  last  terms,  it  is  required  to 
insert  i  arithmetical  means  between  them.  Find  the  expres- 
sion for  the  i  terms  required. 


GEOMETRICAL    PROGRESSION.  207 

CHAPTER     II. 

GEOMETRICAL     PROGRESSION. 

311.  Def.  A  Geometrical  Progression  consists  of 
a  series  of  terms  of  which  each  is  fornied  by  multiply- 
ing the  term  preceding  by  a  constant  factor. 

All  arithmetical  progression  is  formed  by  continual  addi- 
tion or  subtraction;  a  geometrical  progression  by  repeated 
multiplication  or  division. 

Def.  The  factor  by  which  each  term  is  multiplied 
to  form  the  next  one  is  called  the  Common  Ratio. 

The  common  ratio  is  analogous  to  the  common  difference 
in  an  arithmetical  progression. 

In  other  respects  the  same  definitions  apply  to  both. 

EXAM  PLE  s. 

2,     6,     18,     54,     etc., 
is  a  progression  in  which  the  first  term  is  2  and  the  common 
ratio  3. 

2     1      ^      ^      ^      etc 

2'     4'     8^ 

is  a  progression  in  which  the  ratio  is  -• 

+  3,     —  6,     +  12,     —  24,    etc., 
is  a  progression  in  which  the  ratio  is  —  2. 

Note.  A  progression  like  the  second  one  above,  formed  by  dividing 
eadi  term  by.  the  same  divisor  to  obtain  the  next  term,  is  included  in  the 
general  definition,  because  dividing  by  any  number  is  the  same  as  multi- 
plying by  the  reciprocal.  Geometrical  progressions  may  therefore  be 
divided  into  two  classes,  increasing  and  decreasing.  In  the  increasing 
progression  the  common  ratio  is  greater  than  1  and  the  terms  go  on  in- 
creasing ;  in  a  diminishing  progression  the  ratio  is  less  than  unity  and 
the  terms  go  on  diminishing. 

Eem.  In  a  progression  in  which  the  ratio  is  negative,  the 
terms  will  be  alternately  positive  and  negative. 


208  OEOMETRIGAL    PROGRESSION'. 

Def.  A  Geometrical  Mean  between  two  quantities 
is  tlie  square  root  of  their  product. 

EXERCISES. 

Form  five  terms  of  each  of  the  following  geometrical  pro- 
gressions : 

1.  First  term,  1 ;  common  ratio,  5. 

2.  First  term,  7  ;  common  ratio,  —  3. 

3.  First  term,  1 ;  common  ratio,  —  1. 

2  3 

4.  First  term,  - ;  common  ratio,  -• 

4  1 

5.  First  term,  - ;  common  ratio,  -• 

Problems  of  Geometrical  Progression. 

313.  In  a  geometrical  progression,  as  in  an  arithmetical 
one,  there  are  five  quantities!,  any  three  of  which  determine 
the  progression,  and  enable  the  other  two  to  be  found.  They 
are 

a,    the  first  term. 

r,    the  common  ratio. 

Uy    the  number  of  terms. 

I,    the  last  term. 

2,    the  sum  of  the  n  terms. 

The  general  expression  for  the  geometrical  progression 
will  be 

a  J    ar,     ar^^    a7%    etc., 

because  each  of  these  terms  is  formed  by  multiplying  the  pre- 
ceding one  by  r. 

The  same  problems  present  themselves  in  the  two  progres- 
sions.    Those  for  the  geometrical  one  are  as  follows : 

Problem  I.  Given  the  first  term,  the  eommorv  ratio, 
and  the  nunider  of  terms,  to  find  the  last  term. 

The  progression  will  be 

a,     ar,     ar^,     etc. 
We  see  that  the  exponent  of  r  is  less  by  1  than  the  number 
of  the  term,  and  since  it  increases  by  1  for  each  term  added,  it 


m^LIFORg^      GEOMETRICAL   PB0OBE88I0N,  209 

must  remain   less   by  1,  how  many  terms  so  ever  we  take. 
Hence  the  n^^  term  is 

I  =:  ar'^-K  (1) 

Problem  II.     Given  the  last  term,  the  cormnon  ratio, 
and  the  number  of  terms,  to  find  the  first  term. 

The  sokition  is  found  by  dividing  both  members  of  (1)  by 
?'^~S  which  gives 

Problem  III.    Given  the  first  term,  the  last  term,  and 
the  number  of  terms,  to  find  the  common  ratio. 

From  (1)  we  find  r'^~^  =  -• 

Extracting  the  (n  -^  ly^  root  of  each  member,  we  have 

1 


=©" 


[The  sohition  of  Problem  IV  requires  us  to  find  n  from 
equation  (1),  and  belongs  to  a  higher  department  of  Algebra.] 

Problem  V.     To  find  the  sum  of  all  n  terms  of  a  geo- 
metrical progression. 

We  have     !>  =  a  -{-  ar  +  ar^  -\-  etc.  +  ar^~K 

Multiply  both  sides  of  this  equation  by  r.    We  then  have 

rl  =  ai^  +  ar"^  -\-  ar^  -\-  etc +  ar^. 

Now  subtract  the  first  of  these  equations  from  the  second. 
It  is  evident  that,  in  the  second  equation,  each  term  of  the 
second  member  is  equal  to  the  term  of  the  second  member  of 
the  first  equation  which  is  one  place  farther  to  the  right. 
Hence,  when  we  subtract,  all  the  terms  will  cancel  each  other 
except  the  first  of  the  first  equation  and  the  last  of  the  second. 
Illustration.  The  following  is  a  case  in  which  a  =  2,T  =  d,n  —  Q', 
fe'  2  =  2  +  6  +  18  +  54+162  +  486. 

32  r=  6  +  18  +  54  +  162  +  486  +  1458. 
Subtracting,    32-2  =  1458  -  2  =  1456, 

or    22  =  1456,    and    2  r=  728. 
14 


210  GEOMETRICAL   PROGRESSION. 

Returning  to  the  general  problem,  we  have 

(r-l)l.  —  ar^'  —  a   —a  {r^  —  1)  ; 

whence,  2  =  a —  =  a- (4)- 

r  —  1  1  —  r  ^  \' 

It  will  be  most  convenient  to  use  the  first  form  when  r  >  1, 
and  the  second  when  r  <  1. 

By  this  formula  we  are  enabled  to  compute  the  sum  of  the 
terms  of  a  geometrical  progression  without  actually  forming 
all  the  terms  and  adding  them. 

EXERCISES 

3 

1.  Given  3d  term  =z  9,  common  ratio  =  -• 

Find  first  5  terms. 

32  2 

2.  Given  5th  term  =  — ,  common  ratio  =  —  -- 

Find  first  5  terms. 

3.  Given  5th  term  =  x^jj'^,  1st  term  z=  yK 
Find  common  ratio. 

4.  Given  1st  term  =  1,  4th  term  =:  a^. 
Find  common  ratio  and  first  3  terms. 

5.  Given  2d  term  =  m,  common  ratio  :=.  —  m. 
Find  first  4  terms. 

6.  A  farrier  having  told  a  coachman  that  he  would  charge 
him  $3  for  shoeing  his  horse,  the  latter  objected  to  thq  price. 
The  farrier  then  offered  to  take  1  cent  for  the  first  nail,  2  for 
the  second,  4  for  the  third,  and  so  on,  doubling  the  amount 
for  each  nail,  which  offer  the  coachman  accepted.  There  were 
32  nails.  Find  how  much  the  coachman  had  to  pay  for  the 
last  nail,  and  how  much  in  all.     (Compare  §  168,  Rem.) 

7.  Find  the  sum  of  11  terms  of  the  series 

2  +  6  +  18  +  etc., 
in  which  the  first  term  is  2  and  the  common  ratio  3. 

8.  If  the  common  ratio  of  a  progression  is  r,  what  will  be 
the  common  ratio  of  the  progression  formed  by  taking 

I.  Every  alternate  term  of  the  given  progression  ? 
II.  Every  71^^^  term  ? 


GEOMETRICAL   PROGRESSION.  211 

9.  The  same  thing  being  supposed,  what  will  be  the  com- 
mon ratio  of  the  progression  of  which  every  alternate  term  is 
equal  to  every  third  term  of  the  given  progression  ? 

10.  Show  that  if,  in  a  geometrical  progression,  each  term 
be  added  to  or  subtracted  from  that  next  following,  the  sums 
or  remainders  will  form  a  geometrical  progression. 

11.  Show  that  if  the  arithmetical  and  geometrical  means 
of  two  quantities  be  given,  the  quantities  themselves  may  be ' 
found,  and  give  the  expressions  for  them. 

1 2.  The  sum  of  the  first  and  fourth  terms  of  a  progression 
is  to  the  sum  of  the  second  and  third  as  21 :  5.  What  is  the 
common  ratio? 

13.  Express  the  continued  product  of  all  the  terms  of  a 
geometrical  progression  in  terms  of  a,  r,  and  71  ? 

Limit  of  the  Sum  of  a  Progression. 

313.  Theorem.  If  the  common  ratio  in  a  geometri- 
cal progression  is  less  than  unity  (more  exactly,  if  it  is 
contained  between  the  limits  —1  and  +1),  then  there 
will  be  a  certain  quantity  which  the  sum  of  all  the 
terms  can  never  exceed,  no  matter  how  many  terms  we 
take. 

For  example,  the  sum  of  the  progression 

111^ 

^+j  +  g  +  etc., 

in  which  the  common  ratio  is  ~,  can  never  amount  to  1,  no 

matter  how  many  terms  we  take.  To  show  this,  suppose  that 
one  person  owed  another  a  dollar,  and  proceeded  to  pay  him  a 
series  of  fractions  of  a  dollar  in  geometrical  progression, 
namely, 

1111 

r     4'     8'     16'     ^^^* 

When  he  paid  him  the  -  he  would  still  owe  another  -, 
when  he  paid  the  j  he  would  still  owe  another  j?  and  so  on. 


212  GEOMETRICAL   PBOGBESSION. 

That  is,  at  every  payment  he  would  discharge  one-half  the  re- 
maining debt.  Now  there  are  two  propositions  to  be  under- 
stood in  reference  to  this  subject. 

I.  The  entire  debt  ean  never  he  discharged  hy  such 
payments. 

For^  since  the  debt  is  halved  at  every  payment,  if  there  was 
any  payment  which  discharged  the  whole  remaining  debt,  the 
half  of  a  thing  would  be  equal  to  the  whole  of  it,  which  is 
impossible. 

II.  The  debt  can  he  reduced  helow  any  assignable 
limit  by  continuing  to  pay  half  of  it. 

For,  however  small  the  debt  may  be  made,  another  pay- 
ment will  make  it  smaller  by  one-half;  hence  there  is  no 
smallest  amount  below  which  it  cannot  be  reduced. 

These  two  propositions,  wliicli  seem  to  oppose  each  other,  hold  the 
truth  between  them,  as  it  were.  They  constantly  enter  into  the  higher 
mathematics,  and  should  be  well  understood.  We  therefore  present 
another  illustration  of  the  same  subject. 

A  B 

I  -  I  I  I      I       I 

Suppose  AB  to  be  a  line  of  given  length.  Let  us  go  one- 
half  the  distance  from  A  to  B  at  one  step,  one-fourth  at  the 
second,  one-eighth  at  the  third,  etc.  It  is  evident  that,  at  each 
step,  we  go  half  the  distance  which  remains.  Hence  the  two 
principles  just  cited  apply  to  this  case.     That  is, 

1.  We  can  never  reach  B  by  a  series  of  «uch  steps,  because 
we  shall  always  have  a  distance  equal  to  the  last  step  left. 

2.  But  we  can  come  as  near  B  as  we  please,  because  every 
step  carries  us  over  half  the  remaining  distance. 

This  result  is  often  expressed  by  saying  that  we  should  reach  B  by 
taking  an  infinite  number  of  steps.  This  is  a  convenient  form  of  expres- 
sion, and  we  may  sometimes  use  it,  but  it  is  not  logically  exact,  because 
no  conceivable  number  can  be  really  infinite.  The  assumption  that  in- 
finity is  an  algebraic  quantity  often  leads  to  ambiguities  and  difficulties 
in  the  application  of  mathematics. 


GEOMETBIGAL   PBOGRESSION.  213 

Def.  The  Limit  of  the  sum  2  of  a  geometrical 
progression  is  a  quantity  which  s  may  approach  so 
that  its  difference  shall  be  less  than  any  quantity  we 
choose  to  assign,  but  which  s  can  never  reach.    . 

EXAMPLES. 

1.  Unity  is  the  limit  of  the  sum 

2.  The  point  B  in  tlie  preceding  figure  is  the  limit  of  all 
the  steps  that  can  be  taken  in  the  manner  described. 

The  following  principle  will  enable  us  to  find  the  limit  of 
the  sum  of  a  progression : 

314.  Principle.  If  r  <  1,  the  power  r^  can  be  made 
as  small  as  we  please  by  increasing  the  value  of  n^  but 
can  never  be  made  equal  to  0. 

Suppose,  for  instance,  that 

_  3  _  1 

^  ~  4  "~  4' 

Then  every  time  we  multiply  by  r  we  diminish  r^  by 

-  of  its  former  value ;  that  is, 
4 

4  4 


3 

—  ^a  — —   A»3 


4  4^ 

etc.        etc.        etc. 

Now  let  us  again  take  the  expression  for  the  sum  of  a 

series  of  n  terms,  namely, 

1-r^ 

2  =  a- , 

1  —  r 

which  we  may  put  into  the  form 

1  —  r       1  —  r 


214  GEOMETRICAL    PR0OBE8S10N, 

If  r  is  less  than  unity,  we  can,  by  the  principle  just  cited, 
make  the  quantity  r'^  as  small  as  we  please  by  increasing  n 
indefinitely.     From  this  it  follows  that  we  can  also  make  the 

term  -^ r^  as  small  as  we  please. 

1  —  r 

Proof.    Let  us  put,  for  brevity, 
h 


1-r' 
so  that  the  term  under  consideration  is 

If  we  cannot  make  hr'^  as  small  as  we  please,  suppose  5  to 
be  its  smallest  possible  value.     Let  us  divide  s  by  k,  and  put 

^-  k 
No  matter  how  small  s  may  be,  and  how  large  k  may  be, 

T,  or  t,  will  always  be  greater  than  zero.     Hence,  by  the  pre- 
fc 

ceding  principle,  we  can  find  a  value  of  n  so  great  that  y^ 
shall  be  less  than  L     That  is, 

,» <  _. 

Multiplying  both  sides  of  this  inequality  by  k, 

kr'^  <  s. 

That  is,  however  small  we  take  s,  we  can  take  n  so  large 
that  kr"^  shall  be  less  than  s,  and  therefore  s  cannot  be  the 
smallest  value. 

Since  2  =: kr^,  I 

1  —  r 

and  since  we  can  make  kr^  as  small  as  we  please,  it  follows 
Limit  of  2  = 


1  — r 

This  is  sometimes  expressed  by  saying  that  when  r  <  1, 

a  -\-  ar  -{-  ar^  +  ar^  +  etc.,  ad  infinitum  = , 

and  this  is  a  convenient  form  of  expression,  which  will  not  lead 
us  into  error  in  this  case. 


OEOMETBICAL    PROGRESSION.  215 

EXERCISES. 

Haying  giyen  the  progression 

1.1   .1.1  .etc 

of  which  the  limit  is  1,  find  how  many  terms  we  must  take  in 
order  that  the  sum  may  differ  from  1  by  less  than  the  follow- 
ing quantities,  namely : 

Firstly,  .001 ;  secondly,  .000  001 ;  thirdly,  .000  000  001. 

To  do  this,  we  must  find  what  power  of  ^  will  be  less  than  .001, 
what  power  less  than  .000  001,  etc. 

What  are  the  limits  of  the  sums  of  the  following  series : 

1.  Q  +  To  +  Ts  +  ^t^v  ctd  infi^iitum. 

o         o  o 

2.  q  +  Q  +  97  +  ^^C'j  ^^  mfinitum, 

3.  Q  —  02  +  Q3  —  etc.,  ad  infinitum. 

4      42      43 

4.  Q  +  Q2  +  93  +  ^^^'^  ^^  infinitum. 

5.  j-p^  +       ^  ^p  +  (Y+~^p  +  ^^^-^  ^^  infinitum. 

6.  ^-^  -  W^f  "^  W^Y'  ~  ^^^''  ^^  infinitum. 

7.  1 1 5 ^H ;:  —  etc.,  ad  infinitum. 

m      ni?      m?'      m*  "^ 

8.  What  is  that  progression  of  which  the  first  term  is  12 
and  the  limit  of  the  sum  8. 

9.  On  the  line  AB  a  man  starts  from  A  and  goes  to  the 
point  c,  half  way  to  B ;  then  he  re- 
turns  to  d^  half  way  back  to  A ;  then        1        1        1  I 


turns  again  and  goes  half  way  to  c, 
then  back  half  way  to  <?,  and  so  on,  going  at  each  turn  half 
way  to  the  point  from  which  he  last  set  out.  To  what  point 
on  the  line  will  he  continually  approach  ? 


216  COMPOUND    INTEREST.  , 

315.  As  an  interesting  application  of  the  preceding  theory, 
we  may  e:5famine  the  problem  of  finding  the  value  of  a  circu- 
lating decimal.  Such  a  decimal  is  always  equal  to  a  vulgar 
fraction,  which  is  obtained  as  in  the  following  examples : 

1.  What  is  the  value  of  the  decimal 

.373737 ? 

We  find  the  figures  which  form  the  period  to  be  37.  Dividing  the 
decimal  into  periods  of  these  figures,  its  value  is 

_3 j^        37     ,   ^_   ,     . 
100  "^  1002  "^  1003  "^ 

=  ^^(rJo  +  i4-2  +  i4-3  +  ^^4 

The  quantity  in  the  parenthesis  is  a  geometrical  progression,  in  which 

a  —  T-r^ ,  r  =  jjr^  •   The  limit  of  its  sum  is  therefore  —  •    Therefore  the 

37 
value  of  the  decimal  is  ^  • 

This  result  can  be  proved  by  changing  this  vulgar  fraction  to  a 
decimal. 

2.  In  the  case  of  a  decimal  which  has  one  or  more  figures 
before  the  period  commences,  we  cut  these  figures  off,  and 
find  the  value  of  them  and  of  the  circulating  part  separately. 
Thus, 

56363  etc.  =  A  +  j^  +  j^  +  etc. 

__   5  63    /^         1     ,      1      ,     ,    \ 

-  10  +  1000  r  "^  ro"o  +  Ioo~2  +  ^^"^7 

__   5  63     100  __  ^        63   _  558  _  31 

~  To  "^  1000'~99~  —  10  "*■  990  "990  ~  55* 

EXERCISES. 

To  what  vulgar  fractions  are  the  following  circulating  deci- 
mals equal : 

I.     .111111 ?  2.     .2222 ? 

3.  .9999 ?  4.     .09999 ? 

5.     .454545 ?  6.     .2454545 ? 

7.     .108108 ?  8.     72454545 ? 


COMPOUND   INTEBEST.  217 

Compound  Interest. 

316.  When  one  loans  or  invests  money,  collects  the  inter- 
est at  stated  intervals,  and  again  loans  or  invests  this  interest, 
and  so  on,  he  gains  compound  interest. 

Compound  interest  can  always  be  gained  by  one  who  con- 
stantly invests  all  bis  income  derived  from  interest,  provided 
that  he  always  collects  the  interest  when  due,  and  is  able  to 
loan  or  invest  it  at  the  same  rate  as  he  loaned  his  principal. 

Problem  I.  To  find  the  amount  of  p  dollars  for  n 
years,  at  c  per  cent,  compound  interest. 

Solution.  At  the  end  of  one  year  the  interest  will  be 
£--,  which  added  to  the  principal  will  make  p\l  -{•  tf^' 

If  we  put        p  =  -T-.  =  the  rate  of  annual  gain, 

the  amount  at  the  end  of  the  year  will  be  ^  (1  +  p). 

Now  suppose  this  whole  amount  is  put  out  for  another 
year  at  the  same  rate.  The  interest  will  be  p  {1  +  p)  p,  which 
added  to  the  new  principal  p{l  +  p)  will  make  jt?  (1  +  p)^. 

It  is  evident  that,  in  general,  supposing  the  whole  sum 
kept  at  interest,  the  total  amount  of  the  investment  will  be 
multiplied  by  1  +  p  each  year.  "  Hence  the  amounfc  at  the  ends 
of  successive  years  will  be 

^;(l  +  p),    p{l-\-py,    p{l-^p)%    etc. 

At  the  end  of  n  years  the  amount  will  be 

^  (1  +  P)^ 

Problem  II.  A  person  puts  out  2>  dollars  every  year, 
letting  the  whole  sum  constantly  accumulate  at  com- 
pound interest,  WTiat  will  the  amount  be  at  the  end  of 
n  years? 

Solution.  The  first  investment  will  have  been  out  at 
interest  n  years,  the  second  n  —  1  years,  the  third  n  —  2  years, 
and  so  on  to  the  7i^^,  which  will  have  been  out  1  year.  Hence, 
from  the  last  formula,  the  amounts  will  be : 


218  COMPOUND   INTEREST. 

Amount  of  1st  payment^    i^  (1  +  pY- 


2d 

a 

p(i  +  pY-\ 

3d 

Si 

p{i  +  pY-\ 

4th 

i( 

p(l+  p)»-K 

5th 

a 

p(l  +  p)'^-*. 

etc. 

etc. 

The  sum  of  the  amounts  is : 

p{l-\'P)  +  p{l  +  py  +  P{1  +  pY  -h  , . ,  .p(l+p)^. 

This  is  a  geometrical  progression,  of  which  the  first  term  is 
p  (1  +  p),  the  common  ratio  1  +  p,  and  the  number  of  terms  w. 
So  in  the  formula  (4),  §  212,  we  put  J)(l+p)  for  a,  1  +  p  for 
r,  and  thus  find, 

2  =p{i4.o)  (l+i^-i  =^(l±£)!!L-^(i±p). 

^"^'^l  +  p  —  1  ^  p 

EXERCISES. 

I.  A  man  insures  his  hfe  for  $5000  at  the  age  of  30,  pays 
for  his  insurance  a  premium  of  80  dollars  a  year  for  32  years, 
and  dies  at  the  age  of  62,  immediately  before  the  33d  payment 
would  have  been  due.  If  the  company  gains  4  per  cent,  inter- 
est on  all  its  money,  how  much  does  it  gain  or  lose  by  the 


insurance 


p 


Note.  Computations  of  this  class  can  be  made  with  great  facility  by 
the  aid  of  a  table  of  logarithms. 

2.  What  is  the  present  value  of  a  dollars  due  n  years  hence, 
interest  being  reckoned  at  c  per  cent.  ? 

Note.     If  p  be  the  present  value,  Problem  I  gives  the  equation, 

3.  What  is  the  present  value  of  3  annual  payments,  of  a 
dollars  each,  to  be  made  in  one,  two,  and  three  years,  interest 
being  reckoned  at  5  per  cent.  ? 

4.  What  is  the  present  value  of  n  annual  payments,  of  a 
dollars  each,  the  first  being  due  in  one  year,  if  the  rate  of  in- 
terest is  c  per  cent.  ?  What  would  it  be  if  the  first  payment 
were  due  immediately  ? 


SECOND  PART. 
ADVANCED    COURSE. 


f 


BOOK    VIII. 

RELATIONS   BETWEEN  ALGEBRAIC 
QUANTITIES. 


Of  Algebraic  Functions. 

317.  Bef.  When  one  quantity  depends  npon  an- 
other in  such  a  way  that  a  change  in  the  value  of  the 
one  produces  a  change  in  the  value  of  the  other,  the 
latter  is  called  a  Function  of  the  former. 

This  is  a  more  general  definition  of  the  word  "  function  "  than  that 
given  in  §  49. 

Examples.  The  time  required  to  perform  a  journey  is  a 
function  of  the  distance  because,  other  things  being  equal,  it 
varies  with  the  distance. 

The  cost  of  a  package  of  tea  is  a  function  of  its  weight,  be- 
cause the  greater  the  weight  the  greater  the  cost. 

An  algebraic  expression  containing  any  symbol  is  a  func- 
tion of  that  symbol,  because  by  giving  different  values  to  the 
symbol  we  shall  obtain  different  values  for  the  expression. 

Bef.  An  Algebraic  Function  is  one  in  which  the 
relations  of  the  quantities  is  expressed  by  means  of  an 
algebraic  equation. 

Example.  If  in  a  journey  we  call  t  the  time,  8  the  average 
speed,  and  d  the  distance  to  be  travelled,  the  relation  between 
these  quantities  may  be  expressed  by  the  equation, 

d  =  st. 

Any  one  of  these  quantities  is  a  function  of  the  other  two, 
defined  by  means  of  this  equation. 

An  algebraic  function  generally  contains  more  than  one 


222  FUNCTIONS. 

letter,  and  therefore  depends  upon  several  quantities.  But  we 
may  consider  it  a  function  of  any  one  of  these  quantities,  se- 
lected at  pleasure,  by  supposing  all  the  other  quantities  to 
remain  constant  and  only  this  one  to  vary.  For  example,  the 
time  required  for  a  train  to  run  between  two  points  is  a  func- 
tion not  only  of  their  distance  apart,  but  of  the  speed  of  the 
train.  The  speed  being  supposed  constant,  the  time  will  be 
greater  the  greater  the  distance.  The  distance  being  constant, 
the  time  will  be  greater  the  less  the  speed. 

Def.  The  quantities  between  which  the  relation  ex- 
pressed by  a  function  exists  are  called  Variables. 

This  term  is  used  because  such  quantities  may  vary  in  value,  as  in 
the  preceding  examples. 

Def.  An  Independent  Variable  is  one  to  which  we 
may  assign  values  at  pleasure. 

The  function  is  a  dependent  variable,  the  value  of  which  is 
determined  by  the  value  assigned  to  the  independent  variable. 

Def,  A  Constant  is  a  quantity  which  we  suppose 
not  to  vary. 

Eem.  This  division  of  quantities  into  constant  and  varia- 
ble is  merely  a  supposed,  not  a  real  one  ;  we  can,  in  an  algebraic 
expression,  suppose  any  quantities  we  please  to  remain  constant 
and  any  we  please  to  vary.  The  former  are  then,  for  the  time 
being,  constants,  and  the  latter  variables. 

iLLUSTRATIOiq".      If  We  put 

d,  the  distance  from  New  York  to  Chicago  ; 
s,  the  average  speed  of  a  train  between  the  two  cities ; 
t,  the  time  required  for  the  train  to  perform  the  jour- 
ney, 

then,  if  a  manager  computes  the  different  values  of  the  time  t 
corresponding  to  all  values  of  the  speed  5,  he  regards  6?  as  a 
constant,  s  as  an  independent  variable,  and  ^  as  a  function  of  5. 
If  he  computes  how  fast  the  train  must  run  to  perform  the 
journey  in  different  given  times,  he  regards  t  as  the  independ- 
ent variable,  and  5  as  a  function  of  t. 


FUNCTIONS.  223 

When  we  have  any  equation  between  two  variables,  wo 
may  regard  either  of  them  as  an  independent  variable  and  the 
other  as  a  function. 

Example.    From  the  equation 
ax  -\-hy  =1  c, 

we  derive 


by 

c 

X 

a 
ax 

+ 

a' 
c 

y 

— 

■""T 

+ 

v 

in  one  of  which  x  is  expressed  as  a  function  of  y^  and  in  the 
other  ^  as  a  function  of  x, 

218.  Names  are  given  to  particular  classes  of  functions, 
among  which  the  following  are  the  most  common. 

1.  Def,  A  Linear  Function  of  several  variables  is 
one  in  which  each  term  contains  one  of  the  variables, 
and  one  only,  as  a  simple  factor. 

Example.     The  expression 

Ax  ^  By  -\-  Cz 

is  a  linear  function  of  x^  y,  and  z,  when  A,  B,  and  C  are  quan- 
tities which  do  not  contain  these  variables. 

A  linear  function  differs  from  a  function  of  the  first  degree 
(§  52)  in  having  no  term  not  multiplied  by  one  of  the  varia- 
bles.    For  example,  the  expression 

Ax  +  By  +  C 

is  a  function  of  x  and  y  of  the  first  degree,  but  not  a  linear 
function. 

The  fundamental  property  of  a  linear  function  is  this: 

If  all  the  variables  he  -multiplied  hy  a  coimnoii  fac- 
tor,  the  function  will  he  multiplied  hy  the  same  factor. 

Proof.  Let  Ax  -}-  By  -\-  Oz  be  the  linear  function,  and  r 
the  factor.  Multiplying  each  of  the  variables  x,  y,  and  z  by 
this  factor,  the  function  will  become 

Arx  -{-  Bry  +  Orz, 
which  is  equal  to         r  {Ax  +  By  +  Cz). 


224  FUNCTIONS. 

Moreover,  a  linear  function  is  the  only  one  which  pos- 
sesses this  property. 

2.  Def.  A  Homogeneous  Function  of  several  va- 
riables is  one  in  which  each  term  is  of  the  same  degree 
in  the  variables.     (Compare  §  52.) 

Example.  The  expression  ax^+'bx^y-\-cyh-\-dz^  is  homo- 
geneous and  of  the  third  degree  in  the  variables  x,  y,  and  z. 

Kem.  a  linear  function  is  a  homogeneous  function  of  the 
first  degree. 

Fui^DAMEKTAL   PkOPERTY  OF  HOMOGENEOUS  FUKCTIOi^S. 

//  all  the  variables  he  multiplied  hy  a  common  factor, 
any  homogeneous  function  of  the  ii*'^  degree  in  those  va- 
riables will  be  multiplied  by  the  n*^  power  of  that  factor. 

Proof,  If  we  take  a  homogeneous  function  and  put  rx  for 
.T,  ry  for  y,  rz  for  z,  etc.,  then,  because  each  term  contains  x, 
y,  or  z,  etc.,  n  times  in  all  as  a  factor,  it  will  contain  r  n  times 
after  the  substitution  is  made,  and  so  will  be  multiplied  by  r^. 

3.  Def.  A  Rational  Fraction  is  the  quotient  of  tvro 
entire  functions  of  the  same  variable. 

A  rational  fraction  is  of  the  form, 

a  -\-'bx  -^  cx^  -\-  etc. 


m  -\-  nx  -\-  px^  4-  etc. 

Any  rational  function  of  a  variable  may  be  expressed  as  a 
rational  fraction.     Compare  §  180. 

Equations  of  the  First  T^e^ee  between  Two 
Variables. 

319.  Since  we  may  assign  to  an  independent  variable  any 
values  we  please,  we  may  suppose  it  to  increase  or  decrease  by 
regular  steps.  The  difference  between  two  values  is  then 
called  an  increment.     That  is, 

Def.  An  Increment  is  a  quantity  added  to  one 
value  of  a  variable  to  obtain  another  value. 


INCREMENTS,  225 

Eem.  If  we  diminish  the  variable,  the  increment  is 
negative. 

Theorem,  In  a  function  of  the  first  degree,  eqnal  in- 
crements of  the  independent  variable  cause  equal  incre- 
ments of  the  function. 

Example.  Let  x  be  an  independent  variable,  and  call  u 
the  function  -^x'^  11,  so  that  we  have 

u  =  -x  +  11. 

If  we  give  x  the  successive  values  —2,  — 1,  0,  1,  2,  etc., 
and  find  the  corresponding  values  of  the  function  u,  they 
will  be 

Values  of  ic,  —  2,  —  1,  0,  1,  2,  3,  4,  etc. 
"       «  u,         8,         9^,     11,     12|,  14,     154,  ^"^y     etc. 

We  see  that,  the  increments  of  x  being  all  unity,  those  of 
y  are  all  1^, 

General  Proof.  Let  au  -\-  hx  =  c  he  any  equation  of  the 
first  degree  between  the  variable  x  and  the  function  u.  Solving 
this  equation  we  shall  have 

c  —  bx       c       b 

u  = = x. 

a  a      a 

Let  us  assign  to  x  the  successive  values, 
r,     r  -\-  h,    r  -\-  27^,     etc., 

the  increment  being  h  in  each  case.    The  correspondnig  values 
of  the  function  u  will  be 

c       d  c       h         h  ^        c       I         %b  -.        , 

r, r n, r /^,     etc., 

a      a         a      a        a  a      a         a 

of  which  each  is  less  than  the  preceding  by  the  same  amount, 

~h.     Hence  the  increment  of  u  is  always h,  which  proves 

the  theorem. 

230.  Geometric  Construction  of  a  Relation  of  the  First 
Degree.    The  relation  between  a  variable  x  and  a  function  u 
of  this  variable  may  be  Shown  to  the  eye  in  the  following  way  .• 
15 


226 


aEOMETRIO   C0N8TBUCTI0M. 


\ 


el'*] 


-X 


-3        -8 


^^gi     y 


Take  a  base  line  AX,  mark  a  zero  point  upon  it,  and  from 
this  zero  point  lay  off  any  values  of  x  we  please.  Then  at  each 
point  of  the  line  corresponding  to  a  value  of  x  erect  a  vertical 
line  equal  to  the  corresponding  value  of  %i.  If  u  is  positive,  the 
value  is  measured  upward ;  if  negative,  downward.  The  line 
drawn  through  the  ends  of  these  values  of  u  will  show,  by  the 
distance  of  each  of  its  points  from  the  base  line  AX,  the  values 
of  u  corresponding  to  all  values  of  x. 

Let  us  take,  as  an  example,  the  equation 


hu  ■\-dx  =  10, 


3 


the  solution  of  which  gives     u  =  2  —  -x. 

o 

Computing  the  values  of  u  corresponding  to  values  of  x 

from  —3  to  +6,  we  find  : 

X  =  -3,     -2,     -1,      0,   +1,      +2,    +3,  +4,   +5,   +6. 
u=  +3f,   +3i       2|,    2,       If,       f,       i,  -I,  -1,  ^1|. 

Laying  off  these  values  in  the  way  just  described,  we  have 
the  above  figure.  Wherever  we  choose  to  erect  a  value  of  u, 
it  will  end  in  the  dotted  line. 

We  note  that  by  the  property  of  functions  of  the  first  de- 
gree just  proved,  each  value  of  u  is  less  (shorter)  than  the  pre- 

ceding  one  by  the  same  amount ;  in  the  present  case  by  -•    It 

is  known  from  geometry  that  in  this  case  the  dotted  line 
through  the  ends  of  u  will  be  a  straight  line. 

We  call  this  line  through  the  ends  of  u  the  equation  line. 


OF  EQUATIONS   OF   THE  FIB8T  DEGBEE,         227 

331,  When  we  can  once  draw  this  straight  line,  we  can 
find  the  value  of  u  corresponding  to  every  value  of  x  without 
using  the  equation.  We  have  only  to  take  the  point  in  the 
base  line  corresponding  to  any  value  of  x^  and  by  measuring 
the  distance  to  the  line,  we  shall  have  the  corresponding  value 
of  u, 

Now  it  is  an  axiom  of  geometry  that  one  straight  line,  and 
only  one,  can  be  drawn  between  any  two  points.  Therefore, 
to  form  any  relation  of  the  first  degree  we  please  between  x 
and  u,  we  may  take  any  two  values  of  x,  assign  to  them  any 
two  values  of  u  we  please,  plot  these  two  pair  of  values  of  u  in 
a  diagram,  draw  the  equation  line  through  them,  and  then 
measure  off,  by  this  line,  as  many  more  values  of  u  as  we 
please. 

Example.  Let  it  be  required  that  for  x=.  -\-l  we  shall 
have  u^=  -\-ly  and  for  a;  =  +5,  «^  =  +  3.  What  will  be  the 
values  of  w  corresponding  to  a;  =r  —  3,  —2,  —1,  0,  etc. 

Drawing  the  base  line  AX  below,  we  lay  off  from  1  the  ver- 
tical line  +1  in  length,  and  from  the  point  5  the  vertical  line 
+  3.  Then  drawing  the  dotted  line  through  the  ends,  we 
measure  off  different  values  of  u,  as  follows: 

X  =  -3,  ^2,  -1,      0,  +1,  +2,     +3,  +4,     +5,  +6,    etc. 
u  =  -1,  -},      0,  +1       1,  +li,  +2,  +21-,  +3,  +3i,  etc. 


EXE  RCISES. 

1.  Plot  the  equation  2u  -{-  dx  —  6. 

2.  Plot  a  line  such  that 

for    X  =  —  6    we  shall  have    u  =  -f  4, 
for    a;  rr:  +  6         "         "  u  =  —4:, 

and  find  the  values  of  u  for  ^  ==  1,  2,  3,  4,  and  5, 


228  GEOMETBIG   G0N8TBUGT10N 

333.  The  algebraic  problem  corresponding  to  the  con- 
struction of  §  220  is  the  following: 

Having  given  two  values  of  y  corresponding  to  two 
given  values  of  x,  it  is  required  to  construct  an  equation 
of  the  first  degree  such  that  these  two  pairs  of  values 
shall  satisfy  it. 

Example  of  Solution,  Let  the  requirement  be  that  of  the 
equation  plotted  in  the  preceding  example,  namely, 

for    X  ^=1  1     we  must  have     t^  —  1, 
for     X  —  b        "  "  u  —  Z, 

The  problem  then  is  to  find  such  values  of  a,  d,  and  c,  that 
in  the  equation 

ax  4-  hit  =  6',  (1) 

we  shall  have  u  =^1  for  x  '=:  1,  and  u  =  3  for  x  =z  5.     Sub- 
stituting these  two  pairs  of  values,  we  find  that  we  must  have 

axl  +  bxl  =  c, 
ax5  +  bxS  =  c; 

or  a  -{-  b  =:  c, 

5a  -\-  dh  =  c. 

Here  a,  ^,  and  c  are  the  unknown  quantities  whose  values 
are  to  be  found,  and  as  we  have  only  two  equations,  we  cannot 
find  them  all.     Let  us  therefore  find  a  and  h  in  terms  of  c. 

Multiplying  the  first  equation  by  3,  and  subtracting  the 
product  from  the  second,  we  have 

2«  =  —  2c    or    a  =i  —  c. 

Multiplying  the  first  equation  by  5,  and  subtracting  the 
second  from  the  product,  we  have 

2Z>  =  4c    or     l  =  2c. 

Substituting  these  vahies  of  a  and  b  in  (1),  we  find  the  re- 
quired equation  to  be 

2c2i  —  ex  =  c. 

We  may  divide  all  the  terms  of  this  equation  by  c  (§  120, 
Ax.  Ill),  giving 

2u  —  X  =  1, 


i 


OF  EQUATIONS   OF   THE  FIRST  DEGBEE.         221) 

thus  showing  that  there  is  no  need  of  using  c.     The  solution 
of  this  equation  gives 

from  which,  for  x  =  —3,  —2,  —1,  etc.,  we  shall  find  the  same 
values  of  u  which  we  found  from  the  diagram. 

EXERCISES. 

Write  equations  between  x  and  y  which  shall  be  satisfied 
by  the  following  pairs  of  values  of  x  and  y. 

1.  For  X  =  2,  y  =r  1 ;  and  for  x  ==  5,  ?/  =:  —  1, 

2.  For  X  =^  —  2,  y  =:  —  1  ;  and  for  x  =  -\-2,  y  =z  +1. 

3.  For  X  =z  —  5,  y  =:  +  2  ;  and  for  x  =:  -{-5,  y  =  — 2. 

4.  For  X  .=  0,  y  =  —  7  ;  and  for  x  =  lb,  y  =  0. 

5.  For  X  z=  26,  y  =  2 ;  and  for  x  =  30,  y  =  3. 

333.  Geometric  Solution  of  Ttvo  Equations  with  Two  Un- 
hnown  Quantities,  The  solution  of  two  equations  with  two 
unknown  quantities  consists  in  finding  that  one  pair  of  values 
which  will  satisfy  both  equations.  If  we  lay  off  on  the  base 
line  the  required  value  of  x,  the  two  values  of  y  corresponding 
to  this  value  of  a:  in  the  two  equations  must  be  the  same  ;  that 
is,  the  two  equation  lines  must  cross  each  other  at  the 
point  thus  found.     Hence  the  following  geometric  solution : 

I.  Plot  the  two  equations  froin  the  same  hase  line  and 
zero  point. 

II.  Continue  the  equation  lines,  if  necessary,  until 
they  intersect. 

III.  The  distance  of  the  point  of  intersection  froin  the 
hase  line  is  the  value  of  y  which  satisfies  both  equations. 

IV.  The  distance  of  the  foot  of  the  y  line  from,  the 
zero  point  is  the  required  value  of  oc. 

EXERCISES. 

Solve  the  following  equations  by  geometric  construction : 

1.  X  —  2u  =  3,     2x  -\-  u  =  6. 

2.  2u  4-  '^.'^  =  4,     Su  -f  X  —  1. 


230  NOTATION   OF  FUNCTIONS. 

334.  Geometric  Explanation  of  Equivalent  and  Inconsist- 
ent Equations,  If  we  have  two  equivalent  equations  (§  200), 
each  value  of  a:  will  give  the  same  value  of  the  other  quantity 
u  or  y.  Hence  the  two  lines  representing  the  equation  will 
coincide  and  no  definite  point  of  intersection  can  be  fixed. 
If  the  two  equations 

au  ■{•  hx  =:  c, 

a'u  +  i'x  =z  c\ 

are  inconsistent  we  shall  have  (§  142), 

I       V  \ 

a       a'  ] 

If  h  be  any  increment  of  x,  the  increments  of  u  in  the  two 

equations    (§219)  will   be A  and ,h   Therefore   these 

^  ^*^        ^  a  a 

increments  will  be  equal,  and  the  two  equation  lines  will  be 
parallel.     Hence, 

To  inconsistent  equations  correspond  parallel  lines, 
ivhich  have  no  point  of  intersection. 

If  the  two  equations  are  equivalent  (§  141,  143),  their  lines 
will  coincide. 


I 


Notation  of  Functions.  ^ 

335.  In  Algebra  we  use  symbols  to  express  any  numbers 
whatever.  In  the  higher  Algebra,  this  system  is  extended 
thus  : 

We  may  use  any  synvbol,  having  a  letter  attached'  to 
it,  to  express  a  function  of  the  quantity  represented  hy 
that  letter. 

Example.  If  we  have  an  algebraic  expression  containing 
a  quantity  x,  which  we  consider  as  a  function  of  x,  but  do  not 
wish  to  write  in  full,  we  may  call  it 

F{x),     or    </>  {x),     or     \x\     or     Ax, 
or,  in  fine,  any  expression  we  please  which  shall  contain  the 
symbol  x,  and  shall  not  be  mistaken  for  any  other  expression. 

In  the  first  two  of  the  above  expressions,  the  letter  x  is  enclosed  in 
parentheses,  in  order  that  the  expression  may  not  be  mistaken  for  x  mul- 
tiplied by  F,  or  0.  The  parentheses  may  be  omitted  when  the  reader 
knows  that  multiplication  is  not  meant. 


NOTATION  OF  FUNCTIONS,  231 

The  fundamental  principle  of  the  functional  notation  is 
this: 

When  a  syiyihol  with  a  letter  attached  represents  a 
function,  then,  if  ive  substitute  any  other  quantity  for 
the  letter  attached,  the  combination  will  represent  the 
function  found  by  substituting  that  other  quantity. 

Example.  Let  us  consider  the  expression  ax^  +  5  as  a 
function  of  Xy  and  let  us  call  it  (j)(x),  so  that 

0  {x)  =  ax^  +  Z>. 

Then,  to  form  (p  (y),  we  write  y  in  place  of  x^  obtaining 

0  (y)  =  af  +  ^. 

To  form  (p  {x  -}-  y),  we  write  x-\-y  in  place  of  x,  obtaining 

(t){x-\-y)  =  a{x  -\-  yY  +  h. 

To  form  (p  (a),  we  write  a  instead  of  x,  obtaining 

0  (a)  =:  a^  -{-  b» 

To  form  0  {ay^)-,  we  put  ay^  in  place  of  x,  obtaining 

0  {ay^)  =  a  {ay^f  +  &  =  ay  +  b. 

The  equation  (p  {z)  z=zO  will  mean 

az^  +d  =  0. 

EXERCISES. 

Suppose  0  {x)  =  ax^  —  a^x,  and  thence  form  the  values  of 

I.     0(^).  2.     0(^).  3.     0(%). 

4.     0  (a;  +  ^).  5.     0  (.^'  +  05).  6.     0  (.T  —  «). 

7.     0  (ic  +  «?/).  8.     (j){x  —  ay).  9.     0  (:i^2)^ 

Suppose  F  (.^')  =  0;^=^,  and  thence  form  the  values  of 

10.     FQj).  II.     F{^),  12.     Fm). 

13.     i^(:^  +  ^).  14.     F{x-y).  is^     F  {1). 

Suppose  /  {x)  =z  x%  and  thence  form  the  values  of 

16.      /(I).  17.      f{x^).  18.      /(.T«). 

19.    /(a^).  20.    /(a,-=).  21.    /(a;«). 


232  FUNCTIONS   OF  SEVERAL    VARIABLES. 

2  2.  Prove  that  if  we  put  </>  (x)  =  a^,  we  shall  have 
0  (a;  +  ^)  =  </>  {x)  X  (/>  Cv),  0  (:r/y)  =  [0  {x)\y  =  [0  (y)]^. 

Let  lis  put  0  (m)  =.  m{m  —  1)  (m —  2)  (m  —  3) ;  thence 
form  the  values  of 

23.     0(6).  24.     0(5).  25,     0(4). 

26.     0(3).  27.     0(2).  28.     0(1). 

29.     0(0).  30.     0(-l).  31.     0(-^)' 

Functions  of  Several  Variables. 

336.  An  algebraic  expression  containing  several 
quantities  may  be  represented  by  any  symbol  having 
tlie  letters  v^hich  represent  the  quantities  attached. 

Examples.     AYe  may  put 

0 {x,  y)  =z  ax—  hy, 

the  comma  heing  inserted  between   x  and  y,  so  that  their 

product  shall  not  be  understood.     We  shall  then  have, 

0  {m,  7i)  =  am  —  hi, 

0  {y,  '^)  =^  cty  —  bx, 

the  letters  being  simply  interchanged. 

(f>{x-{-y,  x—y)  z=i  a{x  +  y)—l{x  —  y) 
—  (a—  b)  X  +  («5  4-  Z>) y. 
(P{a,b)  =  a^-P. 
0  {d,  a)  =  ab  —  da  =  0. 
^{a  -\-  b,  ab)  =  a  {a  -\-  b)  —  aP. 
0  (a,  a)  =  a^  —  ba, 
etc.  etc. 

If  we  put  0  (^,  b,  c)  =  2^  +  3Z>  —  5c,  we  shall  have 
0  (:r,  z,  y)  =^  ^x  -\-  ?fZ  —  by, 
0  {z,  y,  x)  :=  2z  +  3y  —  6x. 
(f)  (?/?.,  m,  —  m)  —  2m  -i-  dm  -{-  6m  =  10m. 
0(3,8,  6)  =  2'd  +  3.8  — 5-6  =  0. 

EXERCISES. 

Let  us  put  0  (x,  y)  z=  3x  —  4:y, 

fix,  y)  =  ax  -^  by, 
f{x,  y,  z)  =  ax  -i-  by  —  abz. 


USE   OF  INDICES,  233 

Thence  form  the  expressions : 


I.     <p  {y,  x).             2.     <t>  {a,  h). 

3- 

«/>(3,  4). 

4.     <?(4,  3).              5.     </.(10,l). 

6. 

/(«,  ^i). 

7.    f(b,a).              8.    f{y,x). 

9- 

/(7,  -3). 

lo.    f{q,  -p).         II.    f{z,x,y). 

12. 

/(*,  a,  3). 

13.    f{a,b,c).         14.    f{a\1^.c^). 

15.    /(— «,   —5,   —  «5). 

T   ,            ,      ,        X        m(m  —  1)  (w  - 

Let  us  put     (m,  n)  =  — -. -^ — 

'■         ^    '    '         n(ti—  l){n  - 

-2) 

-3) 

Find  the  values  of 

16.     (3,3).                17.     (4,3). 

18. 

(5,  3). 

19.     (6,3).                20.     (7,3). 

21. 

(8,  3). 

22.     (2,  -1).            23.     (3,  -2). 

24. 

(4,  -2). 

Use  of  Indices. 

2SOa.  Any  number  of  different  quantities  may  be 
represented  by  a  common  symbol,  the  distinction  being 
made  by  attaching  numbers  or  accents  to  the  symbol. 

EXAMPLES. 

1.  Any  n  different  quantities  may  be  represented  by  the 
symbols,  p^,  p^,  p^, pn. 

2.  A  producer  desires  to  have  -an  algebraic  symbol  for  the 
amount  of  money  which  he  earns  on  each  day  of  the  year.  If 
he  calls  q  what  he  earns  in  a  day  he  may  put : 

q^     for  the  amount  earned  on  January    1, 

etc.       "  "  "  "     etc., 

^3  3       "  "  ''      February  1; 

and  so  on  to  the  end  of  the  year,  when 

^3  6  5  ^^^1  ^6  ^^^  amount  for  December  31. 

Def.  The  distinguishing  numbers  1,  2,  3,  etc.,  are 
here  called  Indices. 

A  symbol  with  an  index  attached  may  represent  a 
function  of  the  index,  as  in  the  functional  notation. 


234  VSE   OF  INDICES. 

EXERCISES. 

Let  us  put  at  ^=^  t{t  +  1).    Then  find  the  value  of 

1.  ^0+^1+^2+ +  <^io- 

2.  Prove  the  following  equations  by  computing  both  mem- 
bers : 


«, 

+ 

^2 

= 

4 

+ 

«2 

+ 

«3 

= 

5 
3«3. 

+ 

«3 

+ 

Ci 

= 

6 

If  we  put  /Si  =  1  +  2  +  3  . .  .  .  +  ^,  we  shall  have 

/Sg  =  1  +  2  =  3. 

>S3z=l  +  2  +  3i=:6,  etc.,  etc. 

Using  the  preceding  notation,  find  the  values  of  the  ex- 
pressions : 

3-     S^-\-  S^  +  Sq  +  8^.  4.     ct^ -\- a^  +  a^  -\-  a^, 

5.     2/S'5  — ag.  6.     2/S'g-^6. 

337.  Sometimes  the  relations  between  quantities  distin- 
guished by  indices  are  represented  by  equations  of  the  first 
degree.    The  following  are  examples : 
Let  us  have  a  series  of  quantities, 

Aq,    A^,    Aq}    A^,    A^,    etc., 
connected  by  the  general  relation, 

Am  =  ^i  +  ^i-i'  W 

It  is  required  to  express  them  in  terms  of  ^^  and  A^, 

We  put,  in  succession,  ^  =  1,  i  =  2,  i  =  3,  etc.     Then, 
when  i  =  1,  we  have  from  (a), 

A^  —  A^  +Aq. 

When  i  =  2,        A.^  —  A^  +  A^  =  2A^  +    ^o- 

i  =  3,        A^  =  A^-{-  Aq  =  3A^  +  2^0- 

i  =  4,        A^  =:  A^  +  A^  ^  6A,  +  3^0- 

I  =  5,         A^  =  A^  +  A^  z=  SA^  +  6Aq, 

and  so  on  indefinitely. 


MISCELLANEOUS  FUNCTIONS,  236 

EXERCISES. 

1.  If  ^ifi  =  Ai  —  Ai^i, 

what  will  be  the  values  of  ^3 . . . .  ^lo,  and  in  what  way  may 
all  subsequent  values  be  determined  ? 

2.  If  Ai^i  =  2Ai  —  Aq, 
find  A2  to  Aq  in  terms  of  ^^  and  A^, 

3.  If  Ai^t  =  iAi  H-  Ai_\,    find  A^  to  ^5. 

4.  If  Ai  =  Ai_\  +  Ji, 

find  the  sum  A^  -{-  A^  -\-  A^  +  ,  ,  .  .  +  An,  in  terms  of  A^^ 
n  and  n.     (Comp.  §  209,  Prob.  V.) 

5.  If  ^zHi  =  rAi, 

find  ^1  +  ^3  +  ^3  +  .  .  .  .  +  ^n?  in  terms  of  J^,  and  n 

6.  If  ^e+i  =  ^*^-4i  +  Ai^\, 
find  ^3,  ^3, ....  ^g,  in  terms  of  ^^  and  A^, 

Miscellaneous  Functions  of  Numbers. 

238.  We  present,  as  interesting  exercises,  certain  elemen- 
tary forms  of  algebraic  notation  much  used  in  Mathematics, 
and  which  will  be  employed  in  the  present  work. 

1.  When  we  have  a  series  of  synabols  the  number 
of  which  is  either  indeterminate  or  too  great  to  be  all 
written  out,  w^e  may  write  only  the  first  tw^o  or  three 
and  the  last,  the  omitted  ones  being  represented  by  a 
row  of  dots. 

Examples.  a,  h,  c,  .  .  .  .  t, 

J.,  /^,  (5,  .  .  •  •  /vO, 
1,  2,  .  .  .  .  7^, 
n  being  in  the  last  case  any  number  greater  than  2. 

The  number  of  omitted  symbols  is  entirely  arbitrary. 

EXERCISES. 

How  many  omitted  expressions  are  represented  by  the  dots 
in  the  following  series: 

I.     1,  2,  3, n.  2.     1,  2,  3, n  —  2. 


236  MISCELLANEOUS  FUNCTIONS 

3.  1,  2,  3, n  +  2. 

4.  n,  n  —  1,  71  —  2, ,  .  ,  ,  n  —  s, 

5.  n,  n  —  I5  7t  —  2, ,  ,  ,  ,  ?i  —  s  —  1. 

6.  n,  n  —  1,  n  —  2,  ,  ,  .  .  n  —  s  -{-1. 

What  will  be  the  last  term  in  the  series : 

7.  2,  3,  4,  etc.,  to  71  terms. 

8.  n,  n  —  1,  n  —  2,  etc.,  to  s  terms. 

9.  2,  4,  6,  etc.,  to  Jc  terms. 

2.  Product  of  the  First  n  Numbers.    The  symbol 

n\ 
is  used  to  express  the  product  of  the  first  n  numberSj 
1-2.3 n. 

Thus,  1 !  =  1. 

2!  =  1-2  =3  2. 

3!  =  1.2.3  =  6. 

4!  =  1.2.3.4  =  24. 

etc.  etc. 

It  will  he  seen  that   2 !  =  2. 1 ! 

3!  =  3.2! 
And,  in  general,        ?^ !  =  ^  (^  —  1) ! 
whatever  number  n  may  represent. 

EXERCISES. 

Compute  the  values  of 

I.     5!  2.     6!  Z'     ^^- 

_7j_  8! 

^*     3!  4!  ^*     3!  5! 

6.  Prove  the  equation  2.4.6.8  ....  27z  :=;  2'»w! 

7 .  Prove  that,  when  n  is  even, 

^  ^  _  ^  (^  —  2)  (7^  —  4) .  .  .  .•  4. 2 

3.  Binomial  Coefficients,    The  binomial  coefficient 

n(ri  —  1)(7^  —  2) .  .  .  .  to  5  terms 

123 s 

is  expressed  in  the  abbreviated  form, 


MISCELLANEOUS  FUNCTIONS,  237 


©■ 


the  parentheses  being  used  to  show  that  what  is  meant 

is  not  the  fraction  -• 
s 


( 


EXAMPLES. 

(?) 

=  ?  =  s. 

© 

7-6.5.4.3 
■~  1.2.3.4.5  ~ 

21. 

(f) 

n 

(1) 

n  (n  —  l){n- 
~            1-2-3 

-2) 

© 

n{n  —  l).. 

..2.1_j 

~       1.3-3... 

.    ^ 

'»  +  4\ 

_  (re  +  4)  {71  +  3)  (w  +  2) 

1.2.3 


EXERCISES. 

Compute  the  vaUies  of  the  expressions : 

-  ©+(^(i)-©-©+(i)-(^©- 

'■  (l)-e)-(^©+(D- 

Prove  the  formulae : 


/5\  _     5!  (n\  __  nl 

^*     \2/  ~"  2TF!  "^^     \s/  ~  si  {n  -  s) ! 

5.  (•^^D  =  !^:(;)-    ^-  ©  +  (l)  =  m' 

'■  (i)+(i)=m- 
«•  (i)+(i)=m- 


BOOK    IX. 
THE     THEORY     OF    NUMBERS. 


CHAPTER    I. 
THE     DIVISIBILITY     OF     NUMBERS. 

329.  Def,  The  Theory  of  Numbers  is  a  branch 
of  mathematics  which  treats  of  the  properties  of  integers. 

Def.  An  Integer  is  any  whole  number,  positive  or 
negative. 

In  the  theory  of  numbers  the  word  nwnber  is  used  to  ex- 
press an  integer. 

Def.  A  Prime  Number  is  one  which  has  no  divi- 
sor except  itself  and  unity. 

The  series  of  prime  numbers  are 

2,  3,  5,  7,  11,  13,  17,  19,  23,  29,  etc. 

Def  A  Composite  Number  is  one  which  may  be 
expressed  as  a  product  of  two  or  more  factors,  all 
greater  than  unity. 

Kem.  Every  number  greater  than  1  must  be  either  prime 
or  composite. 

Def.  Two  numbers  are  prime  to  each  other  when 
they  have  no  common  divisor  greater  than  unity. 

Example.  The  numbers  24  and  35  are  prime  to  each 
other,  though  neither  of  them  is  a  prime  number. 

Eem.  a  vulgar  fraction  is  reduced  to  its  lowest  terms  when 
numerator  and  denominator  are  prime  to  each  other. 


DIVISIBILITY   OF  NUMBERS,  239 

Division  into  Prime  Factors. 

330.  Every  composite  number  may  by  definition  be  di- 
vided into  two  or  more  factors.  If  any  of  these  factors  are 
composite,  they  may  be  again  divided  into  other  factors. 
When  none  of  the  factors  can  be  farther  divided,  they  will  all 
be  prime.    Hence, 

Theokem.  Every  composite  number  may  he  divided 
into  prime  factors. 

Example.  180  rr  9-20, 

9  z=  3.3, 
20  =  4.5  =  2.2.5. 
Whence,  180  =  2-2.3.3.5  =  22.32.5. 

Cor,  1.  Because  every  number,  not  prime  is  composite, 
and  because  every  composite  number  may  be  divided  into 
prime  factors,  we  conclude:  Every  nitrnber  is  either  prime 
or  divisible  hy  a  prime. 

Cor,  2.  Every  number,  prime  or  composite,  may  be  ex- 
pressed in  the  form 

p°-q^ry  etc.,  (a) 

where  ^,  q,  r,  etc.,  are  different  prime  numbers ; 

S  ft  y^  ^tc,  the  exponents,  are  positive  integers. 

Rem.  If  the  number  is  prime  there  will  be  but  one  factor, 
namely,  the  number  itself,  and  the  exponent  will  be  unity. 

EXERCISES. 

Divide  the  following  numbers  or  products  into  their  prime 

factors,  if  any,  and  thus  express  the  numbers  in  the  form  {a)  : 

I.     24.         2.     72.         3.     260.         4.     169.         5.     225. 

6.     256.       7.     91.         8.     143.         9.     360.       10.     217. 

II.     3072.  12.     1.2.3.4.5.6.7.8.9. 

kEEM.  In  seeking  for  the  prime  factors  of  a  number,  it  is 
lever  necessary  to  try  divisors  greater  than  its  square  root,  for 
f  a  number  is  divisible  into  two  factors,  one  of  these  factors 


240  DIVISIBILITY   OF  NUMBERS, 

Coinmon  Divisors  of  Two  Numbers, 

331.  Theoeem  I.  //  two  mnnhers  have  a  common 
factor,  tJieir  sum  will  have  that  same  factor. 

Proof,    Let     a  be  the  common  factor  ; 

m,  the  product  of  all  the  other  factors  in  the 

one  number; 
n^  the  corresponding  product  in  the  other 
number. 
Then  the  two  numbers  will  be 

am.    and    an. 
Their  sum  will  be  a  (m  +  ^). 

Because  m  and  n  are  whole  numbers,  m-^n  will  also  be  a 
whole  number.     Therefore  a  will  be  a  factor  of  am  +  an. 

Theoeem  II.  //  two  numhers  have  a  comm>on  factor, 
their  difference  will  have  the  same  factor. 

Proof.     Almost  the  same  as  in  the  last  theorem. 

Cor,  If  a  number  is  divisible  by  a  factor,  all  multiples  will 
be  divisible  by  that  factor. 

Eem.     The  preceding  theorems  may  be  expressed  as  follows : 

//  two  nioinhers  are  divisible  hy  the  saiiie  divisor, 
their  sum,  difference,  and  multiples  are  all  divisihle  hy 
that  divisor, 

Eem.  If  one  number  is  not  exactly  divisible  by  another,  a 
remainder  less  than  the  divisor  will  be  left  over.    If  we  put 

Z),  the  dividend; 
d,  the  divisor; 
q,  the  quotient; 
r,  the  remainder; 

we  shall  have,  D  :=  dq  -\-  r, 

or  D  —  dq  =:  r. 

Example.    7  goes  into  QQ  9  times  and  3  over.     Heno 
this  means 

66  =  7.9  +  3,     or    66 --7.9  =  3. 


I 


DIVISIBILITY  OF  NUMBERS.  241 

333.  Peoblem.  To  find  the  greatest  common  divisor 
of  two  numbers. 

Let  m  and  n  be  the  numbers,  and  let  m  be  the  greater. 

1.  Divide  m  by  n.  If  the  remainder  is  zero,  n  will  be  the 
divisor  required,  because  every  number  divides  itself.  If  there 
IS  a  remainder,  let  q  be  the  quotient  and  r  the  remainder. 

Then  m  —  nq  =  r. 

Let  d  be  the  common  divisor  required. 
Because  m  and  n,  are  both  divisible  by  d,  m  —  nq  must 
also  be  divisible  by  d  (Theorem  II).     Therefore, 

r  is  divisible  by  d. 

Hence  every  common  divisor  of  m  and  n  is  also  a  common 
divisor  of  n  and  r.     Conversely,  because 
m  =z  nq  +  r, 

every  common  divisor  of  n  and  r  is  also  a  divisor  of  m.  There- 
fore, the  greatest  common  divisor  of  m  and  n  is  the  same  as 
the  greatest  common  divisor  of  /^  and  r,  and  we  proceed  with 
these  last  two  numbers  as  we  did  with  7)1  and  n, 

2.  Let  r  go  into  n  q'  times  with  the  remainder  r'. 
Then  n  =  rq  -\-  r\ 

or  n  ^rq'  ^=  r'. 

Then  it  can  be  shown  as  before  that  6?  is  a  divisor  of  r\  and 
therefore  the  greatest  common  divisor  of  r  and  r'. 

3.  Dividing  r  by  r\  and  continuing  the  process,  one  of  two 
results  must  follow.     Either, 

cc.  We  at  length  reach  a  remainder  1,  in  which  case  the 
two  numbers  are  prime  ;  or, 

P,  We  have  a  remainder  which  exactly  divides  the  pre- 
ceding divisor,  in  which  case  this  remainder  is  the  divisor 
required. 

To  clearly  exhibit  the  process,  we  express  the  numbers  m, 
n,  and  the  successive  remainders  in  the  following  form  : 
16 


242  GREATEST   COMMON  DIVISOR, 

m  =  n-q  +  r,  {r  <  n); 

n  =  r-q'  +  r',  (r'  <  r); 

r  =  r'-q"  -f->",  (/'  <  r'); 

r'  =  r"'q"'  +  r"\  {r'"  <  r") -, 

etc.  etc.  etc., 

iinfcil  we  reach  a  remainder  equal  to  1  or  0,  when  the  series 
terminates. 

EXERCISES. 

I.  Find  the  G.  C.  D.*  of  240  and  155. 


ividend 

. 

Div. 

Quo.    Rem. 

240 

r= 

155 

.1  +  85. 

155 

Z=i 

85 

•1  +  70. 

85 

Z=l 

70 

1  +  15. 

70 

=: 

15 

4  +  10. 

15 

= 

10 

1+    5. 

10 

— 

5 

2. 

Therefore  5  is  the  greatest  common  divisor. 

Note.  Let  the  student  arranf^e  all  the  following  exercises  in  the 
above  form,  first  dividing  in  the  usual  way,  if  he  finds  it  necessary. 

Find  the  greatest  common  divisor  of 

2.  399  and  427.  3.  91  and  131. 

4.  8  and  13.  5.  1000  and  212. 

6.  799  and  1232.  7.  800  and  1729. 

8.  250  and  625.  9.  1000  and  370. 

10.  If  p  be  a  number  less  than  n  and  prime  to  n,  show  that 
n  —p  is  also  prime  to  n, 

11.  It  p  be  any  number  less  than  ?i,  the  greatest  common 
divisor  between  7i  and  p  is  the  same  as  that  between  n  and 
n  —-p. 

12.  If  7z  is  any  odd  number,  — ~ —  and  — 7^ —  are  both 
prime  to  it. 

Corollaries,  1.  When  two  numbers  are  divided  by  their 
greatest  common  divisor,  their  quotients  will  be  prime  to  each 
other. 

*  The  letters  G.  C.  D.  are  an  abbreviation  for  Greatest  Common  Divisor. 


GEARING    OF   WHEELS. 


243 


I 


2.  Conversely,  if  two  numbers,  n  and  n\  prime  to  each 
other,  are  each  multiplied  by  any  number  d,  then  d  will  be  the 
G.  C.  D.  of  dn  and  dn', 

233.  Gearing  of  Wheels.  An  interesting  problem  con- 
nected with  the  greatest  com- 
mon divisor  is  afforded  by  a 
common  pair  of  gear  wheels. 
Let  there  be  two  wheels,  the 
one  having  m  teeth  and  the 
other  n  teeth,  gearing  into  each 
other.  If  we  start  the  wheels 
with  a  certain  tooth  of  the  one 
against  a  certain  tooth  of  the 
other,  then  we  have  the  questions : 

(1. )  How  many  revolutions  must  each  wheel  make  before 
the  same  teeth  will  again  come  together  ? 

(2. )  With  how  many  teeth  of  the  one  will  each  tooth  of  the 
other  have  geared  ? 

Let  q  be  the  required  number  of  turns  of  the  first  wheel, 
having  m  teeth. 

Let^  be  the  required  number  of  turns  of  the  second,  hav- 
ing n  teeth. 

Then,  because  the  first  wheel  has  m  teeth,  qni  teeth  will 
have  geared  into  the  other  wheel  during  the  q  turns.  In  the 
same  way,  pn  teeth  of  the  second  wheel  will  have  geared  into 
the  first.  But  these  numbers  must  be  equal.  Therefore, 
when  the  two  teeth  again  meet, 

pn  =  qm. 

Conversely,  for  every  pair  of  numbers  of  revolutions  2^  and 
q,  which  fulfil  the  conditions, 

pn  =.  qm, 

the  same  teeth  will  come  together,  because  each  wheel  will 
have  made  an  entire  number  of  revolutions.  This  equation 
gives 

p m 

q~  n 


244  QEABING    OF   WHEELS, 

Hence,  if  we  reduce  the  fraction  —  to  its  lowest  terms,  we 

n  '  ^ 

shall  have  the  smallest  number  of  revolutions  of  the  respective 
wheels  which  will  bring  the  teeth  together  again. 

To  answer  the  second  question  : 

After  the  first  wheel  has  made  q  revolutions,  qm  of  its  teeth 
have  passed  a  fixed  point.  Any  one  tooth  of  tlie  other  wheel 
gears  into  every  n^  passing  tooth  of  the  first  wheel.    Therefore 

QTYl 

any  such  tooth  has  geared  into  - —  teeth  of  the  first  wheel, 

that  is,  into  p  teeth,  because,  from  the  last  equation, 

qm 

^—  —p. 
n 

If  d  be  the  G.  0.  D.  of  m  and  n,  then 

m  =  dp^ 

n  :=  dq\ 

m 

n 

Therefore  each  tooth  of  the  one  wheel  has  geared  into  only 
every  d^  tooth  of  the  other. 

In  the  figure  on  the  preceding  page,  m  =  21  and  /^  =  6. 
Hence,  ^  =  3,  and  each  tooth  of  the  one  will  gear  into  every 
third  tooth  of  the  other.  The  numbers  on  the  large  wheel 
show  the  order  in  which  the  gearing  occurs. 

How  long  soever  the  wheels  run,  the  same  contacts  will 
be  repeated  in  regular  order.  Hence,  if  each  tooth  of  the 
one  wheel  must  gear  with  every  tooth  of  the  other,  the 
numbers  m  and  n  must  he  prime  to  each  other. 

EXERCI  SES. 

1.  If  one  wheel  has  40  teeth  and  the  other  10,  show  how 
they  will  run  together. 

Show  the  same  thing  for  the  following  cases: 

2.  w?  r=  72,  w  =  15.  3.     m  =1  24:,  n  =  18. 
4.     m  =  36,  71  =  25.  5;    m  =  24,  7i  =  7. 


1 


NUMBERS  AND    THEIR   DIGITS.  245 

Relations  of  Numbers  to  their  Digits. 

334.  In  our  ordinary  method  of  expressing  numbers,  the 
second  digit  toward  the  right  expresses  lO's,  the  third  lOO's, 
etc.  That  is,  each  digit  expresses  a  power  of  10  correspond- 
ing to  its  position. 

Def.  The  number  10  is  the  Base  of  our  scale  of 
numeration. 

ISToTE.  The  base  10  is  entirely  arbitrary,  and  is  supposed 
to  have  originated  from  the  number  of  the  thumbs  and  fingers, 
these  being  used  by  primitive  people  in  counting. 

Any  other  number  might  equally  well  have  been  chosen  as 
a  base,  but  in  any  case  we  should  need  a  number  of  separate 
characters  (digits)  equal  to  the  base,  and  no  more. 

Had  8  been  the  base,  we  should  have  needed  only  the 
digits  0,  1,  2,  etc.,  to  7,  and  different  combinations  of  the 
digits  would  have  represented  numbers  as  follows : 
1  =  1, 
7  =  7, 
10  =  1-8  +  0  =  eight. 
17  =  1-8  +  7  =  fifteen. 
20  =  2-8  +  0  =  sixteen. 
56  =  5*8  -f  6  r=  forty-six. 
234  =  2-82  +  3-8  +  4  =  one  hundred  fifty-six, 
etc. 
Let  us  take  the  arbitrary  number  z  as  the  base  of  the  scale. 
As  in  our  scale  of  lO's  we  have 

234  =  2.102  +  3.10  +  4, 

so  in  the  scale  of  ^'s  the  digits  234  would  mean 

2^2  +  3;^  +  4. 

In  general,  the  combination  of  digits  abed  would  mean 

az^  +  dz^  +  cz  -\-  d. 

Divisibility  of  Numbers  and  tlieir  Digits, 

235.  Theorem.    If  the  sum  of  the  digits  of  any  num 
her  he  subtracted  from  the  nurriber  itself,  the  remainder 
will  he  divisible  by  z  —  1. 


246  DIVISIBILITY   OF  NUMBERS.  J 

Proof.  Let  the  digits  be  a,  b,  c,  d.  The  number  expressed 
wiU  be  az^  j^  M  j^  cz  +  d 

Sum  of  digits  =:  a     -\- h    -{-  c   -\-  d 

Subtracting,    rem.  =  a{z^—l)  +  h(f—l)  +  c{z—l). 

The  factors  z^  —  1,  z^  —  1,  and  z  —  1  are  all  divisible  by 
2;  —  1  (§  93).     Hence  the  theorem  is  proved.     (§  231.) 

Theorem.  Ijv  any  scale  having  z  as  its  base,  the  sum 
of  the  digits  of  any  number,  when  divided  by  z  —  1,  ivill 
leave  the  same  remainder  as  will  the  number  itself  when 
so  divided. 

If  we  put :      n,  the  number ;    s,  the  sum  of  the  digits  ; 
r,  r',  the  remainders  from  dividing  by  z—lj 
q,  q' ,  the  quotients  ;    we  shall  have, 
^Number,  n  =1  q{z  —  1)  -\-  r 

Sum  of  digits,     s  =  q'  {z  —  1)  +  r' 
Eemainder,  {q  —  q')  {z  —  1)  -\-  r  —  r. 

Because  7i  —  s  and  {q  —  q')  {z  —  1)  are  both  divisible  by 
z  —  1,  their  difference  r  —  r'  must  be  so  divisible.  Since  r 
and  r'  are  both  less  than  z  —  1,  this  remainder  can  be  divided 
by  ;2;  —  1  only  when  r  =  r',  which  proves  the  theorem. 

Zero  is  considered  divisible  by  all  numbers,  because  a  re- 
mainder 0  is  always  left. 

If  a  be  any  factor  oi  z  —  1,  the  same  reasoning  will  apply 
to  it,  and  therefore  the  theorem  will  be  true  of  it. 

In  our  system  of  notation,  where  z  =  10,  the  above  theo- 
rems may  be  put  in  the  following  well-known  form : 

If  the  sunt'  of  the  digits  of  any  number  be  divisible 
by  3  or  9,  the  number  itself  ivill  be  so  divisible. 

These  are  the  only  numbers  of  which  the  theorem  is  true, 
because  3  is  the  only  divisor  of  9. 

Theorem.  If  from  any  number  we  subtract  the  digits 
of  the  even  powers  of  z,  and  add  those  of  the  alternate 
-powers,  the  result  will  be  divisible  by  z  -\-  1. 

Proof.     To  az^  +  bz^  +  cz  +  d 

Add         a    —  b    -\-  c  —  d 


Result,    a  (f^-\)  ^b{z^—l)  +  c(z^\). 


NUMBERS   AND    THEIR   DIGITS,  247 

The  factors  of  a^  h,  and  c  are  all  divisible  by  2;  +  l  (§§  93, 
94),  whence  the  result  itself  is  so  divisible. 

Applying  this  result  to  the  case  of  ^  =  10,  we  conclude: 

//  on  suhtracting  the  sum  of  the  digits  in  the  place 
of  units,  hundreds^  tens  of  thousands,  etc.,  froin  the  sum 
of  the  alternate  ones,  the  remainder  is  divisible  hy  11, 
the  number  itself  is  divisible  by  11. 

If  771  be  any  factor  of  ^,  it  will  divide  all  the  terms  of  the 
number 

az^  +  bz^  -\-  c^  +  d, 

except  the  last.-  Hence,  if  it  divide  this  last  also,  it  will  di- 
vide the  number  itself.  Applying  this  result  to  the  case  of 
z  =  10,  we  conclude  : 

If  the  last  digit  of  any  number  is  divisible  by  a  fac- 
tor of  10,  the  number  itself  is  divisible  by  that  factor. 

The  factors  of  10  being  2  and  5,  this  rule  is  true  of  these 
numbers  only. 

It  will  be  remarked  that  if  the  base  of  the  system  had  been 
an  odd  number,  we  could  not  have  distinguished  even  and  odd 
numbers  by  their  last  figure,  as  we  habitually  do. 

For  example,  if  the  base  had  been  9,  the  figures  72  would 
have  represented  what  we  call  sixty-five,  which  is  odd,  and  73 
would  have  represented  what  we  call  sixty-six,  which  is  even. 

The  use  of  the  base  10  makes  it  easy  to  detect  when  a  num- 
ber is  divisible  by  either  of  the  first  three  prime  numbers,  2,  3, 
and  5.  If  the  last  figure  is  divisible  by  2  or  5,  the  whole  num- 
ber is  so  divisible.  To  ascertain  whether  3  is  a  factor,  we  find 
whether  the  sum  of  the  digits  is  divisible  by  3. 

In  taking  the  sum,  it  is  not  necessary  to  include  all  tlie  digits,  but  in 
adding  we  may  omit  all  3's  and  9*s,  and  drop  3,  6,  or  9  from  the  sum  as 
often  as  convenient.     Thus,  if  the  number  were 

921642712, 
we  should  perform  the  operation  mentally,  thus : 

Drop  9  ;  3  +  1  =  3,  which  drop  ;  6,  drop  ;  4  +  2  =  6,  which  drop; 
7  +  1  =  8  +  2  =  10,  which  leaves  a  remainder  1. 

EXERCISES. 

T.  Prove  that  if  an  even  number  leaves  a  remainder  1  when 
divided  by  3,  its  half  will  leave  a  remainder  2  when  so  divided. 


248  DIVISIBILITY   OF  NUMBERS. 

2.  If  from  any  number  we  subtract  the  sum  of  units'  digit 
plus  the  product  of  the  tens'  digit  by  ^,  plus  the  product  of 
the  hundreds'  digit  by  i%  etc.,  the  remainder  will  be  diyisible 
by  10  —  i.     (^  may  be  any  integer,  positive  or  negative.) 

Note.  When  i  —  1,  this  gives  the  rule  of  9's  and  when  ^  =  —  1,  the 
rule  of  ll's. 

Prime  Factors  of  Numbers. 

336.  First  Fui^damental  Theorem.  A  product  can- 
not he  divided  hy  a  prime  numher  unless  one  of  the  fac- 
tors is  divisible  by  that  prime  number. 

Note.  This  theorem  is  not  true  of  composite  divisors.  For  exam- 
ple, neither  8  nor  9  is  divisible  by  6,  but  the  product  8  •  9  =  72  is  so 
divisible.  But  if  we  take  as  many  numbers  as  we  please  not  divisible  by 
7,  we  shall  always  find  their  product  to  leave  a  remainder  when  we  try 
to  divide  it  by  7. 

To  make  the  demonstration  better  understood,  we  shall  first  take  a 
special  case : 

The  product  Q^a  is  not  divisible  by  7,  unless  a  is  divisible 
byl. 

Proof.     Suppose .    .    .    .     66^  di  v.  by  7 

7  goes  into  66  9  times  and  3  over,  because  7  •  9  =  63,  63a  di v.  by  7 

Therefore,  by  Theorem  II,  §  231, 3a  div.  byl' 

2 

3  goes  into  7  2  times  and  1  over.     Multiply  by  2,     6a  div.  by  7 

Subtracting, 7a  div.  by  7 

We  have  left, a  div.  by  7 

Hence,  if  66a  is  divisible  by  7,  then  a  is  divisible  by  7. 

Gauss's  Demonstration,  If  it  be  possible,  let  am  be  the 
smallest  multiple  of  m  which  is  divisible  by  p,  when  neither  a 
nor  m  is  so  divisible.  If  a  is  greater  than  p,  then  let  p  go 
into  a  b  times  and  r  over,  so  that 

a  :=  bp  +  r, 
or  a  —  bp  z=z  r. 

Then,  am  div.  by  p. 

Subtract  bpm  "        " 

Eemainder,  (a  —  bp)  m  "        " 

Or  r77i  "        " 


PRIME  FACTORS   OF  NUMBERS.  249 

That  is,  if  am.  is  divisible  by  p,  so  is  ririy  where  r  is  less 
than  p. 

Therefore  the  smallest  multiple  of  m  which  fulfils  the  con- 
ditions must  be  less  than  pm. 

Therefore,  let  a  <i  p.     Let  a  go  into  p  c  times  and  s  over, 
so  that 

p  z=  ca  -\-  s, 
or  p  -^  ca  =z  Si 

Then  pm  div.  by  p, 

cam    "        "     (by  hypothesis). 
Subtracting,         (p  —  cd)  m    "        " 
Or,  sm    "        " 

Therefore,  8  being  less  than  «,  a  is  not  the  smallest  multiple; 
whence  the  hypothesis  that  a  is  the  smallest  is  impossible. 

General  Demonstration,     Suppose 

jt?,  a  prime  number  ; 
«,  number  not  divisible  by^; 
am,  a  product  divisible  by  p. 

We  have  to  prove  that  m  must  be  divisible  by  p, 
Letp  go  into  a  q  times.     Because  a  is  not  divisible  byj9, 
a  remainder  r  will  be  left.     That  is, 

a  =:  j)q  -\-  T,     or     a  —  pq  =r  r. 

Let  r  go  into  p  q'  times  and  leave 
a  remainder  r'.     Then, 

p  =  q'r  +  r', 

and  because  ;?m  and  q'rm  are  both  di- 
visible by^,  rm  is  so  divisible. 

In  the  same  way,  if  r'  goes  into  p 
q"  times,  and  leave  the  remainder  r'\ 
r"m  will  be  divisible  by^.  Since  each 
of  the  remainders  r,  r',  r",  etc.,  must  r"m 

be  less  than  the  preceding,  we  shall  at 
length  reach  a  leniainder  1,  which  will  give 

m  divisible  by  p.        Q.  E.  D. 


am 

div. 

hj  p. 

pqm 

a 

U 

rm 

a 

u 

q'rm 

iC 

iC 

pm 

a 

a 

r'm 

a 

iC 

q'Vm 

a 

(C 

pm 

a 

6i 

250  DIVISIBILITY   OF  NUMBEB8. 

Extension  to  Several  Factors,  If  7n  is  a  product  l  x  n^  and 
d  is  not  divisible  by  p,  then  we  may  show  in  the  same  way  that 
n  must  be  so  divisible.  If  7i  =  cs,  and  c  is  not  divisible,  then 
s  must  be  divisible,  and  so  on  to  any  number  of  factors. 

Hence, 

Theoeem.  If  a  product  of  any  number  of  factors  is 
divisible  by  a  prime  number,  then  one  of  the  factors 
must  be  divisible  by  the  same  prime. 

This  theorem  is  the  logical  equivalent  of  the  one  just 
enunciated  as  the  first  fundamental  theorem. 

Note.  The  student  will  remark  why  the  preceding  demonstration 
applies  only  when  the  divisor  p  is  a  prime  number.  If  it  were  composite, 
we  might  reach  a  remainder  which  would  exactly  divide  it,  and  then  the 
conclusion  would  not  follow. 

237.  Second  Fui^DAME:N^TAL  Theoeem.  A  nimiber 
can  be  divided  into  prime  factors  in  only  one  way. 

For,  suppose  we  could  express  the  number  N  in  the  two 
ways  (§  204,  Cor.  2), 

JSf  =  p"-  q^  ry, 

JV  =  af^b""  C", 
where  jt?,  q,  r,  etc.,  a,  h,  c,  etc.,  are  all  prime  numbers.     Then 

p<^q^ry  =  a^lfC", 

If  common  prime  factors  appeared  on  both  sides  of  this 
equation,  we  could  divide  them  out,  leaving  an  equation  in 
which  the  prime  factors  ^,  ^,  r,  etc.,  are  all  different  from 
a,  b,  c,  etc. 

Then,  because  a,  h,  c,  etc.,  are  all  prime,  none  of  them  are 
divisible  by^.  Therefore,  by  the  first  fundamental  theorem, 
their  products  cannot  be  so  divisible.  But  the  left-hand  mem- 
ber of  the  equation  is  divisible  by  p,  because  p  is  one  of  its 
factors.     Therefore  the  equation  is  impossible. 

Eem.  This  theorem  forms  the  basis  of  the  theory  of  the 
divisibility  of  numbers. 

The  preceding  theorems  enable  us  to  place  the  definition 
of  numbers  prime  to  eacli  other  in  a  new  shape. 


BINOMIAL    COEFFICIENTS.  251 

Two  nnmbers  are  said  to  be  prime  to  each  other 
when  they  have  no  common  prime  factors. 

Example.  If  one  number  is  p'^q^ry,  and  the  other  is 
a^h^'c^  {p,  q,  r,  etc.,  and  a,  h,  c,  etc.,  being  prime  numbers), 
then,  lip,  q,  r,  etc.,  are  all  different  from  a,  h,  c,  etc.,  the  two 
numbers  will  be  prime  to  each  other. 

Elementary  Theorems. 

338.  The  following  general  theorems  follow  from  the  two 
preceding  fundamental  theorems,  and  their  demonstration  is 
in  part  left  as  an  exercise  for  the  student. 

I.  JVo  power  of  an  irreducible  vulgar  fraction  can  he 
a  ivlhole  number,  ■ 

Note.  An  irreducible  vulgar  fraction  is  one  which  is  re- 
duced to  its  lowest  terms. 

II.  CoKOLLAEY.  JVo  root  of  a  whole  number  ca^n  be  a 
vulgar  fraction. 

III.  //  a  number  is  divisible  by  several  divisors,  all 
prime  to  each  other,  it  is  also  divisible  by  their  product. 

Cor,  To  prove  that  a  number  N  is  divisible  by  a  number 
B  ^zp'^q^  ry,  it  is  sufficient  to  prove  that  it  is  divisible  sepa- 
rately by  p%  by  ^^,  by  rv,  etc. 

Example.  If  a  number  is  divisible  separately  by  5,  8,  and 
9,  it  is  divisible  by  5-  8-  9  =  360.  Hence,  to  prove  that  a  num- 
ber is  divisible  by  360,  it  is  sufficient  to  show  that  5,  8,  and  9 
are  all  factors  of  it. 

I       IV.  //  the  numerator  and  denominator  of  a  vulgar 
^  fraction  have  no  common  prime  factors,  it  is  reduced  to 
its  lowest  terms. 


Binomial  Coefficients. 

239.  Theorem.  The  product  of  any  n  consecutive 
numbers  is  divisible  by  the  product  of  the  numbers 
1-2-3  .  .  .  .  ?z,  or  71 ! 


252 


BINOMIAL    COEFFICIENTS. 


Eem.  The  theorem  implies  that  all  binomial  coefficients 
are  whole  numbers,  because  they  are  quotients  formed  by  di- 
viding the  product  of  n  consecutive  numbers  by  n ! 

Proof,  1.  We  have  first  to  find  the  prime  factors  of  the 
product 

1.2.3.4.5.6 n  =  n\ 

beginning  with  the  factor  2. 

I.  The  numbers  divisible  by  2  are  the  even  numbers  2,  4, 

6,  etc.,  to  n  or  n  —  1,  the  number  of  which  is 


Note.     The  expression 


here  means  the  greatest  tvJiole 
n  —  1 


number  in  - ,  which  is  ^  itself  when  7i  is  even,  and 
when  n  is  odd. 

The  quotients  of  the  division  are 
1,  2,  3,  4,  .• .  .  . 


are  divisible  by  2,  leaving  the 


Of  these  quotients, 
second  set  of  quotients, 

1,  2,  3,  ...  . 

The  next  set  of  quotients  will  be 

1,  2, 


The  process  is  to  be  continued  until  we  have  no  even  num- 
bers left. 

Therefore,  if  we  put  a  for  the  number  of  times  that  the 
factor  2  enters  into  n !  we  have, 

+  etc. 

II.  The  numbers  in  the  series  n !  containing  3  as  a  factor  are 
3,  6,  9,  12,  etc., 


n 

2 

+ 

n 
4 

+ 

5] 

BINOMIAL    COEFFICIENTS. 


253 


of  which  the  number  is 
Tiding  them  by  3  are 


The  quotients  obtained  by  di- 


1,  2,  3,  . 


Of  these  quotients, 


[1] 


are  again  divisible  by  3,  and  so 


on  as  before.     Hence,  if  we  put  (i  for  the  number  of  times  n\ 
contains  3  as  a  factor,  we  have 


^  = 


n 

+ 

n 
9 

+ 

71 

+  etc. 


In  the  same  way,  if  k  be  any  prime  number,  n !  will  con- 
tain yb  as  a  factor 

\V\  +  [I]  +  [1]  +  '^-  *™^'- 

Note.  This  elegant  process  enables  us  to  find  all  the  prime 
factors  of  n\  without  actually  computing  it,  and  thus  to  ex- 
hibit n !  as  a  product  of  prime  factors.  If  we  suppose  n  =  12, 
we  shall  find, 

12!  =  1.2-3 12  r=  210.35.52.7.11. 

2.  Next  let  us  find  the  prime  factors  of  the  product 

(^  +  l)(«  +  2) (a  +  n), 

which  contains  n  factors.  Dividing  su^ccessively  by  2,  3,  5,  7, 
etc.,  it  is  shown  in  the  same  way  as  before  that  the  prime  fac- 
tor ji;  is  contained  in  the  product  at  least  , 


+ 


p^ 


-(-  etc.  times, 


whatever  prime  factor  p  may  be.  Therefore  the  numerator 
(a  +  l)  (a  +  2) . . . .  (a  +  n)  contains  all  the  prime  factors  found 
in  n\  to  at  least  the  same  power  with  which  they  enter  nl 
Hence  (§  238,  III),  the  numerator  is  divisible  by  n ! 

Cor.    If  the  factor  ^  +  ^  in  the  numerator  is  a  prime 
number,  that  prime  cannot  be  contained  in  n !  because  it  is 


254  DIVISORS   OF  A    NUMBER. 

greater  than  a.    Hence  the  binomial  factor  will  be  divisible 

by  it. 

5«6«7 
Example.  -  is  divisible  by  7. 

We  may  show  in  the  same  way  that  the  binomial  coefficient 
is  divisible  by  all  the  prime  numbers  in  its  numerator  which 
exceed  n. 

Divisors  of  a  Number. 

340.  Def,    The  expression 

(/)(m) 
is  used  to  express  how  many  numbers  not  greater  than 
m  are  prime  to  m. 

Example.     Let  us  find  the  value  of  0  (9). 

1  is  prime  to  9,  because  their  G.  C.  D.  is  1. 

2  a  a  a  ((  a  a 

3  is  not  prime  to  9,  because  their  G.  C.  D.  is  3. 

4  is  prime  to  9. 

5  "  " 

6  is  not,  because  6  and  9  have  the  G.  C.  D.  3. 

7  is. 

8  is. 

9  is  not. 

Therefore,  the  numbers  less  than  9  and  prime  to  it  are 

1,  2,  4,  5,  7,  8, 

which  are  six  in  number.    Hence, 

0 (9)  ^  6. 

The  numbers  less  than  12  and  prime  to  12  are  1,  5,  7,  11. 
Hence, 

9  (12)  =  4. 

,    We  find  in  this  way, 

0(1)  =1,  0(2)  =  1,  0(3)  =2, 

0(4)  =2,  0(5)  =4,  0(6)  =2, 

0  (7)  —  6,  etc.,                           etc. 


DIVISORS   OF  A    NUMBER.  255 

Cor,  1.  The  number  1  is  prime  to  itself,  but  no  other 
number  is  prime  to  itself. 

Cor.  2,     If  m  be  a  prime  number,  then 
0  [m)  z=z  m  —  1, 
because  the  numbers   1,  2,  3, .  = . .  ^?2  —  1    are  then  all  prime 
to  m. 

The  following  remarkable  theorem  is  associated  with  the 
functions  0  (m). 

241.  Theorem,  If  N  be  any  number,  and  d^^  d^^ 
d^^  etc.,  all  its  divisors,  unity  and iV^ included,  then 

0  {d^)  +  0(^2)  +  ^(^3)  +  etc.  -^  N. 

Example.    Let  the  number  be  18. 

The  divisors  are  1,  2,  3,  6,  9,  18.    We  find,  by  counting, 

.       -  0(1)    -    1 

</>(2)    =    1 

0(3)  =  2 
0(6)  ^  2 
0(9)    =    6 

0  (18)  =  _6 

Sum,     18. 

To  show  how  this  comes  about,  write  down  the  numbers 
1  to  18,  and  under  each  write  the  greatest  common  divisor  of 
that  number  and  18.     Thus, 

Num.,    1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18. 
G.C.D.,  12321612921612321  18. 

Necessarily  the  numbers  in  the  second  line  are  all  divisors 
of  18  as  well  as  of  the  numbers  over  them. 

The  divisor  1  is  under  all  the  numbers  prime  to  18,  so 
that  there  are 

0  (18)  =  divisors  1. 

If  n  be  any  number  over  the  divisor  2,  then  -  and  -^,  or 
9,  must  be  prime  to  each  other.     (§  232,  Cor.  1.)     That  is,  the 


256  DIVISORS  OF  A   NUMBER. 

numbers  n  are  all  those  which,  when  divided  by  2,  are  prime 
to  9.     So  there  are 

(f>  (9)  divisors  2. 

The  divisor  3  marks  all  numbers  which,  when  divided  by  3, 

18 
are  prime  to  —  =  6.     Hence,  there  are 

o 

0  (6)  divisors  3. 

In  the  same  way  there  are  0  (3)  divisors  G,  0  (2)  divisors  9, 
and  (j)  (1)  divisor  18. 

The  total  number  of  these  divisors  is  both  18  and  (p  (18) 
+  </)  (9)  +  etc.     Hence, 

0(18)4-0(9)  +0(6) +0(3) +  0(2)  +  0(1)  =  18. 

General  Proof.    Let  m  be  the  given  number; 
^1,  d^,  ^3,  etc.,  its  divisors; 

0)1-      77Z 

Qiy  ^2^  C's'  ^^^^  quotients  -^,  -^,  etc. 

J  3 

The  quotients  q^,  q^,  etc.,  will  be  the  same  numbers  as  d^, 
c?2,  etc.,  only  in  reverse  order.  The  smallest  of  each  row  will 
be  1  and  the  greatest  m.    We  shall  then  have 

m  =  d^q^  =  d^q^  z=z  d^  q^,    etc. 

From  the  list  of  numbers  1,  2,  3,  ...  .  m,  select  all  those 
which  have  d^  (unity)  as  the  greatest  common  divisor  with  m, 
then  those  which  have  d^  as  such  common  divisor,  then  those 
which  have  d^,  etc.,  till  we  reach  the  last  divisor,  which  will 
be  m  itself,  and  which  will  correspond  to  m. 

The  numbers  having  unity  as  G.  C.  D.  will  be  those  prime 
to  m,  by  definition.     Their  number  is  0  (m). 

Those  having  d^  as  G.  C.  D.  with  7n  will,  when  divided  by 

771 

cZg,  give  quotients  prime  to  -r-  or  to  ^g.    Moreover,  such  quo- 

tients  will  include  all  the  numbers  not  greater  than  ^g  and 
prime  to  it,  because  each  of  these  numbers,  when  multiplied 
by  ^g,  will  give  a  number  not  greater  than  m,  and  having  d^ 
as  its  G.  0.  D.  with  m.     Hence  the  number  of  numbers  not 


FEBMAT8   THEOREM,  257 

greater  than  m,  and  having  d^  as  its  G.  C.  D.  with  m  will 
be  0(^2). 

Continuing  the  process,  we  shall  reach  the  divisor  m^  which 
will  have  m  itself  as  G.  C.  D.,  and  which  will  count  as  the 
number  corresponding  to  0  (1)  =  1  in  the  list. 

The  m  numbers  1,  2,  3, ....  m  are  therefore  equal  in  num- 
ber to 

0(m)  +  0(^,)  +  (/)(^3)+  ..-.  +^(1); 

or,  since  the  quotients  and  divisors  are  the  same,  only  in  re- 
verse order,  we  shall  have 

343.  Fermat's  Theorem.  //  p  he  any  prim^e  num- 
ber, and  a  be  a  number  prime  to  p,  then  a^-^  —  1  will  be 
divisible  by  p. 

Examples.     «*  —  1  is  divisible  by  5 ;  a^  ■—  1  is  divisible  by  7. 

Proof,  Develop  aP  in  the  following  way  by  the  binomial 
theorem, 

«*=[!  +  («-  1)]^ 

=  1  +pia  -1)  +  {l)ia-iy  +  . ..  .  +  {a-  1)". 

Because  p  is  prime,  all  the  binomial  cocflficients. 


p,     (I),    etc.,     to     (^^^y 


are  divisible  by^  (§  239,  Cor.).  Transposing  the  terms  of  the 
last  member  of  the  equation  which  are  not  divisible  hj  p,  we 
find 

a^  —  (a  —  ly  —  1  =  a  multiple  of  p. 

or       a^  ■—  a  —  \{a  —  1)^  —  («^  —  1)]  =  a  multiple  of  p. 

Supposing  a  =  2,  this  equation  shows  that  2^  —  2  is  a 
multiple  of  p ;  then,  supposing  a  =  3,  we  show  by  §  231, 
Th.  II,  that  3^  —  3  is  such  a  multiple,  and  so  on,  indefinitely. 

Hence,  oF  —  «  ==  a  multiple  of  p^ 

whatever  be  a.     But  a^  —  a  =z  {aP~^  —  1)  a,  and  because  this 
product  is  divisible  by  p,  one  of  its  factors  must  be  so  divisible 
(§  236).     Hence,  if  a  is  prime  to  p,  aP~^  —  1  is  divisible  hy  p, 
17 


258  CONTINUED   FRACTIONS. 

CHAPTER     II. 

OF     CONTINUED      FRACTIONS. 

243.  Any  proper  fraction  may  be  represented  in  the  form 
-;- ,  where  x^  is  greater  than  unity,  but  is  not  necessarily  a  whole 
number.    If  a ^  be  the  greatest  whole  number  in  a; j,  we  can  put 

where  x^  will  be  greater  than  unity.     In  the  same  way  we 
may  put 

_  1_ 

X2  —  ^2  +      , 

_  1 

X^    —   ^3    +  ":> 

etc.  etc. 

If  for  each  x  we  substitute  its  expression,  the  fraction  — 
will  take  the  form  ,  * 

1  11 


X.  1  1 


'"^  «3.-j ,  etc.,  etc. 

x^ 

If  the  substitutions  are  continued  indefinitely,  the  form 
will  be  1 


1 

«i  +- 


1 

«2    +- 


1 
«3    +- 


Such  an  expression  is  called  a  continued  fraction. 

Def.  A  Continued  Fraction  is  one  of  which  the 
denominator  is  a  whole  number  plus  a  fraction ;  the 
denominator  of  this  last  fraction  a  whole  number  plus 
a  fraction,  etc. 


CONTINUED    FRACTIONS.  259 

A  continued  fraction  may  either  terminate  with  one  of  its 
denominators  or  it  may  extend  indefinitely. 

Bef.    When  the  number  of  quotients  a  is  finite,  the 
fraction  is  said  to  be  Terminating, 

344.  Problem.     To  find  the  value  of  a  continued 
fraction. 

We  first  find  the  value  when  we  stop  at  the  first  denomina- 
tor, then  at  the  second,  then  at  the  third,  etc.     * 
Using  only  two  denominators,  the  fraction  will  be 

11  Xc> 


F  z=. 


1        a.x^  +  1' 

Xn 


F  being  put  for  the  true  value  of  the  fraction. 

To  find  the  expression  with  three  terms,  we  put,  in  the 

preceding  expression,  a^  ■\ in  place  of  x^.     This  gives 

x^ 

1 

rp ^3  __  ^2^3    +   -*- . 

"     n    n      O-^l    J-   1      ~    (^1^2    +1)^3    +«1* 
^1^3    +  —  +-•- 

To  find  the  result  with  the  fourth  denominator,  we  substi- 
tute x^i=lcu  A The  fraction  becomes: 

3         ^    '  x^ 

^  ^  (^2^3  +1)^:4  +^a ,  X 

\{a^ac^  +  1)  «3  +  ^1]  ^4  +  ^1^2  +1 

To  investigate  the  general  law  according  to  which   the 
successive  expressions  proceed,  we  put 

P,  the  coefficient  of  :c  in  any  numerator; 
P',  the  coefficient  of  x  in  the  denominator ; 
Q^  the  terms  not  multiplied  by  x  in  the  numerator  ; 
§',  the  terms  not  multiplied  by  x  in  the  denominator  ; 

and  we  distinguish  the  various  expressions  by  giving  each  P 
and  Q  the  same  index  as  the  x  to  which  it  belongs. 


260  CONTINUED   FRACTIONS, 

Then  we  may  represent  each  value  of  F  in  the  form, 

where  i  may  take  any  yalue  necessary  to  distinguish  the  frac- 
tion.    Comparing  with  the  fractions  as  written,  we  see  that : 

P,  =0,  Q,  =1,     P^,  =1,  Q\  =0; 

F,  =1,  Q,  =  0,     P;  =  a,,  (?;  -  1 ;       (6) 

To  show  that  this  form  will  continue,  how  far  soever  w^e 
carry  the  computation,  we  put  in  the  expression  (b)  the  general 
value  of  Xi, 

1 

Xi  =z  ai  -i , 

I.-  1,    •  jjy      {ctiPi-\-  Qi)  n+1  +  Pi  f.. 

wh.chg..es,         F  =  ^^p-^-^^^-^r  (,) 

To  show  the  general  law  of  succession  of  the  terms,  let  us 
compare  the  general  equation  (b)  with  (d).  Putting  ^  +  l  for 
i  in  (b),  it  becomes, 

Comparing  this  with  (d),  we  find 

Fi+t  =z  aiFi  -\-  Qi, 

Qibl  =  Pi\ 

whence,  Qi  =  P^^i. 

Substituting  this  value  of  Qi  in  the  equation  previous,  it 
becomes 

P,+i  =  ^^P^  +  Pi_i.  (/) 

Working  in  the  same  w^ay  with  the  denominators,  we  find 

Q'i.t  =  P'v 
By  supposing  i  to  take  in  succession  the  values  1,  2,  3,  etc., 


CONTINUED   FRACTIONS,  26i 

these  formulae  show  that  the  successive  values  of  P  may  be 
computed  thus : 

pUiJ   ^'""'^' 

Pg  =  a,P,  +  P„ 
P,  =  a,P,  +  P^, 
etc.,  to  any  extent. 

Also,  P;  =  1, 

p;  =  «!,  ^ 

.  -P 3    =^   tt^Ps   +  Pi, 

P^    =   ^S-^S    +  -f*2» 

etc.  etc. 

Since  each  value  of  Q  is  equal  to  the  value  of  P  having  the 
next  smaller  index,  it  is  not  necessary  to  compute  the  $'s  sep- 
arately. 

If  the  fraction  terminates  at  the  n^^  value  of  a,  we  shall 

have 

Xn  =  an,  exactly. 

If  it  does  not  terminate,  we  have  to  neglect  all  the  denom- 
inators after  a  certain  point ;  and  calUng  the  last  denominator 
we  use  the  n*^,  we  must  suppose 

Xji  "=.  dn* 

In  either  case,  the  expression  (Z>)  will  give  the  value  of  the 
fraction  with  which  we  stop  by  putting  i  =:n  and  Xn  =  an- 

Therefore,  F  =  "^^r^  ^f, 

Ctn  Pn  +   Qn 

or,  substituting  for  Q'^  and  Q'^  their  values  in  (^), 

■p ^n  I^n  "T"  I^n—1 

But  the  general  expressions  (/)  and  (g)  give 


262  CONTINUED   FRACTIONS.     ' 

(^n  't^n  "f"  ^n—\  "^^^  -^n+h 
(^n  Pn  +  Pn-\  =  Pn+1* 

Therefore,  F=^' 

Pn+l 

Therefore,  to  find  the  value  of  the  fraction  to  the  n^^ 
term,  we  have  only  to  comjjute  the  values  of  Pn+i  and 
Pn+h  without  tahing  any  account  of  Q, 

Example.     Take  the  fraction, 
1 


i  +  i 


2  4-i 


3  +  i 


4  +  ^ 


5  etc. 
Here,    a^  =1,    CI2  =  2,    a^  =  3,  ,  .  ,  .  ai  z=  i. 

We  now  have,  by  continuing  the  formulae  (c)  and  (/),  and 
using  those  values  of  a^,  a 2,  etc. : 

Pi  =0, 
P2  =  1. 

Ps    ^^  ^3-^2    +  Pi    =^2=2^ 

P4    =  «3^3    +   Pg    =   3.2  +  1   =   7, 

P,  =a^P^  +  Pg  =4.7  +  2  =  30, 
Pe  =  a,P,  +Pi  =  5.30  +  7  =  157, 
etc.  etc.  etc. 

p;  =  i, 

P'a  =  «i  =  1, 

p;  =  a,P',  +  P\  =  3-1  +  1  =  3, 

p;  =  (jjp;  +  p;  =  3.3  + 1  =  10, 
p',  =  «4^4  +  K  =  4-10  +  3  =  43,. 

p;  =  a,P',  +  P;  =  5.43  +  10  =  225. 

Therefore,  supposing  in  succession,  n  =  l,  n  —  2,  n  =  d, 
etc.,  we  have,  for  the  successive  approximate  values  of  tlie 
fraction, 


•. 


CONTINUED    FRACTIONS. 

For  n  =  l, 

2 

For  ^  =  2, 

^«       P\  -  3 

For  n  =  6, 

^  ~  P'~  225' 

263 


These  successive  approximate  values  of  the  continued  frac- 
tion are  called  Converging  Fractions,  or  Convergents. 

345.  The  forms  (/)  and  {g)  may  be  expressed  in  words  as 
follows : 

The  numerator  of  each  convergent  is  formed  by  mul- 
tiplying the  preceding  numerator  by  the  corresponding 
a,  and  adding  the  second  numerator  preceding  to  the 
product. 

The  successive  denominators  are  formed  in  the  same  way. 

Example.     The  ratio  of  the  motions  of  the  sun  and  moon 
relative  to  the  moon's  node  is  given  by  the  continued  fraction : 
1 

12  +  -^ 


i  +  i 


2  +  i 


1+^ 


4  +  ^ 


3  +  etc. 

Let  us  find  the  successive  convergents.  We  put  the  de- 
nominators «>t  =  12^  ^2  =  1,  etc.,  in  a  line,  thus: 

a     =    12,     1,       2,       1,       4,        3. 
P     =     0      J^      J^       3        4        19        61 

P'    =     1'     12'     13'     38'     51'     242'     777* 

Under  a^  we  write  the  fraction  -,  whicli  is  always  the  one 
with  which  to  start,  because  P^  =  0  and  P\  =  1  (§  244,  c). 
Next  to  the  right  is  — ,  because  Pg  =  1  and  Pg  =  a.  After 
this,  we  multiply  each  term  by  the  multiplier  a  above  it,  and 


264  CONTINUED   FRACTIONS. 

add  the  term  to  the  left  to  obtain  the  term  on  the  right. 
Thus,     2.1  +  1  =  3,    2.13  +  12  =  38,    etc. 

Ex.  2.    To  compute  the  convergents  of 

1 


^+     1 

^  +  4  etc. 

a    = 

3,     4,     2,      4,      2, 

4, 

Numerators, 
Denominators, 

0      1      4       9       40 

1'     3'     9'     20'     89' 

89 
198' 

etc. 
etc. 


EXERCISES. 


Reduce  the  following  continued  fractions  to  vulgar  frac- 
tions : 


3  +  ^.  3  +  ^ 


^  +  4-  2  +  ^ 


3  +  1 


3 


4- 


1  ^1 

3  +  ——  3  + a  + 


1  .1  '1 

1+ 5  +  -.  b  +  -. 

3  +  ^  "^  '■ 

1  +  r 

346.  Problem.     To  express  a  fraetional  quantity  as 
n  eontinued  fraetion. 

Let  R  be  the  given  fraction,  less  than  unity.     Compute  x^ 
from  the  formula, 

_   1 
^1  -  ^• 

Let  a^  be  the  whole  number  and  R'  the  fraction  of  ^^j. 
Then  compute 

__   1 

^2    -^- 


CONTINUED   FRACTIONS. 


265 


Let  a^  be  the  whole  number  and  R"  the  fraction  of  x^. 
We  continue  this  process  to  any  extent,  unless  some  value 
^f  X  comes  out  a  whole  number,  when  we  stop. 

26 


Example.    Express  ^  as  a  continued  fraction. 


1         73 
^'^  ^    :^   "^  26  ^  ^  +  26  ' 


R' 
Jl_ 
R" 


26 
21 
21 


1  + 


21 
21^ 


5   ~^  +  5^ 


So  the  continued  fraction  is 
1_ 

2  +  1- 


^1  =  2 


ttg  =  1 


«3  =  4 
^4  =  5 


R' 
R" 


21 

26* 

21* 


R'"  —  -. 
R"  =  0. 


It  will  be  seen  that  the  process  is  the  same  as  that  of  find- 
ing the  greatest  common  divisor  of  two  numbers. 


EXERCISES. 


Develop  the  following  quotients  as  continued  fractions : 

113  1049  628 

^'     925* 


355 


2. 


3326 


347.  The  most  simple  continued  fraction  is  that  arising 
from  the  geometric  problem  of  cutting  a  line  in  extreme  and 
mean  ratio.     The  corresponding  numerical  problem  is : 

To  divide  unity  into  two  such  fractions  that  the  less 
shall  he  to  the  greater  as  the  greater  is  to  unity. 

Let  r  be  the  greater  fraction.  Then  1  —r  will  be  the 
lesser  one.    We  must  then  have 

1  —  r  I  r    : :    r  :  1, 


266  CONTINUED   FRACTIONS. 

which  gives  r^  =  1  --  r, 

or  r^  4-  r  =  1, 

or  r  (r  +  1)  =  1, 

1 


or  r  z=z 


1  +  r 

Now,  let  us  put  for  r  in  the  last  denominator  the  expression 
,  and  repeat  the  process  indefinitely.     We  shall  have. 


1  +  ^ 

1 

r  z=z  - 


i  +  i 


i  +  i 


1  +  ^ 


1  etc.,  ad  infiniUim, 

Now  we  may  form  the  successive  convergents  which 
approximate  to  the  true  value  by  the  rule.  As  all  the  denom- 
inators a  are  1,  we  have  no  multiplying,  but  only  add  each 
term  to  the  preceding  one  to  obtain  the  following  one.  Thus 
we  find: 

011235^132134 

1^    1'    2'    3''  5'    8'    13'    21'    34'    55'    ^*''' 

The  true  value  of  r  may  be  found  by  solving  the  quadratic, 

y2  -f  r  =  1, 

.  .  ,     .                                         -1±  a/5 
which  gives  r  = ^r • 

The  positive  root,  with  which  alone  we  are  cOYiCemed,  is 

r  =:  "-^j"  ^^  =  0.61803399. 

The  values  of  the  first  nine  convergents,  with  their  errors. 

are: 

1:1=  1.0,  error  =  +  0.382. 
1:2=  0.5,  "  —0.118. 

2:3=  0.666....,         "  +0.0486. 

3:5=  0.600,  ''  -0.0180. 

5:8=  0.625,  ''  +-0.00697. 


CONTINUED    FRACTIONS.  267 

8  :  13  =  0.61538.... ,  error  =  —  0.00265. 

13  :  21  =  0.61904.... ,  "           +  0.00101. 

21  :  34  =  0.617647.... ,  "          -  0.000397. 

34  :  55  =  0.618182....,  "           +  0.000148. 

etc.               etc.  etc. 

Relations  of  Successive  Convergents. 

348.  Theorem  I.  The  successive  conver gents  are 
alternately  too  large  OMid  too  small. 

Proof,     The  first  convergent  is   —    The  true  denom- 

inator  being  a^  -\ ,  the  denominator  a^  is  too  small,  and 

therefore  the  fraction  is  too  large. 

In  forming  the  second  fraction,  we  put  —  instead  of  — 

a.2  x^ 

Because  ^2  <  i?:^3,  this  fraction  is  too  large,  which  makes  the 
denominator  a^  -\ too  great. 

The  third  denominator  a^  is  too  small,  which  will  make 
the  preceding  one  too  large,  the  next  preceding  too  small,  and 
so  on  alternately. 

171  7Yi! 

Theorem  II.    //  —  and  —,  he  any  two  consecutive 

convergents,  then 

mn!  —  m'n  =  ±  1. 

Proof,    We  show  : 

(«)  Thatthe  theorem  is  true  of  the  first  pair  of  con  vergents. 

{p)  That  if  true  of  any  pair,  it  will  be  true  of  the  pair  next 
following. 

{a)  The  first  pair  of  convergents  are 
J_  1        _        ^2 

""^■^^ 
which  gives  7nn'  —  m'n  =  1,  thus  proving  («). 


268  CONTINUED   FRACTIONS, 

\p)   Let  — ,      — 7,      —77, 

^  ^  n       n        n 

be  three  consecutive  convergents,  in  which 

7nn'  —  m'n  =.  ±1,  (1) 

By  (/)  and  {g)  we  shall  have 

^"  =  an'  +  ^. 
Multiplying  the  second  equation  by  m'  and  subtracting  the 
product  of  the  first  by  n',  we  have       * 

m'n"  —  m"n'  r=  m'71  —  mn', 

Avhich  is  the  negative  of  (1),  showing  that  the  result  is  =F  1- 

The  theorem  being  true  of  the  first  and  second  fractions, 
must  therefore  be  true  of  the  second  and  third  ;  therefore  of 
the  third  and  fourth,  and  so  on  indefinitely. 

Corollaries,     Dividing  (1)  by  n7i\  we  have 

m       w!         .     1         T_r 

;  =  ±  — -,-      Hence, 

n        n  nn 

I.  The  difference  between  the  two  successive  converge 
ents  is  equal  to  unity  divided  hy  the  product  of  the 
denominators. 

Because  the  denominator  of  each  fraction  is  greater  than 
that  of  the  preceding  one,  we  conclude : 

II.  T]^e  difference  between  two  consecutive  convergents 
constantly  diminishes. 

Combining  these  conclusions  with  Th.  T,  we  conclude  : 

III.  Each  value  of  a  convergent  always  lies  between 
the  values  of  the  two  preceding  convergents. 

For  if  R^,  R^,  R^  be  three  such  fractions,  and  if  R^  is 
greater  than  R^,  then  R^  will  be  less  than  R^.  But  it  must 
be  greater  than  R^,  else  we  should  not  have  R^  —  Rq  numer- 
ically less  than  R^  —  R^.  Hence,  if  we  arrange  the  successive 
convergents  in  a  line  in  the  order  of  magnitude,  their  order 
will  be  as  follows : 

-"'4^  -^^e^  -*^8'   •  •  •  •   ^9?  Ri9  R^9 
each  convergent  coming  nearer  a  true  central  value.    Hence, 


CONTINUED    FRACTIONS.  269 

IV.  The  true  value  of  the  continued  fraction  al- 
ways  lies  between  the  values  of  two  consecutive  con- 
vergents. 

Comparing  with  (I),  we  conclude : 

V.  The  error  which  we  mahe  hy  stopping  at  any  con- 
vergent can  never  he  greater  than  unity  divided  hy  the 
product  of  the  denominators  of  that  convergent  and  the 
one  next  following. 

EXAM  PLE. 

Referring  to  the  table  of  values  of  |-(\/5  —  1)  in  §  247, 
we  see  that : 


Error  of    2  : 

^^<3-5' 

(for  .0486  <  1). 

Error  of    3  ; 

^^<5'8' 

(for  .018    <!). 

etc. 

etc. 

Hence,  in  general,  continued  fractions  give  a  very  rapid 
approximation  to  the  true  value  of  a  quantity.  Their  princi- 
pal use  arises  from  their  giving  approximate  values  of  irrational 
numbers  by  vulgar  fractions  with  the  smallest  terms. 

EXAMPLE. 

Develop  the  fractional  part  of  ^2  as  a  continued  fraction, 
and  find  the  values  of  eight  convergents. 

Because  1  is  the  greatest  whole  number  in  V^^  we  put 

V2  =  14-^;  (1) 

whence,  oc 


a/2  — 1 

Eationalizing  the  denominator,  §  185, 
X  =  V2  +  1. 
Substituting  for  V2  its  value  in  (1), 

a;  =  2  +  -. 

X 


270  CONTINUED   FRACTIONS. 

Putting  this  value  of  x  in  (1)  and  again  in  the  denominator, 
and  repeating  the  substitution  indefinitely,  we  find 

V3  =  1  +  i 


.  +  i 


2  +  i 


^  +  2  etc. 

Forming  the  convergents,  we  find  them  to  be 

12       5       12      29       70       169      408 
2'     5'     12'    29'     70'     169'     408'     985' 


etc. 


3 

7 

17 

41 

99 

239 

577 

1393 

3' 

5' 

12' 

29' 

70' 

169' 

408' 

985 

Adding  unity  to  each  of  them,  we  find  the  approximate 
values  of  ^2 : 

17        4.1         QQ        9,?JQ        .^77        I.^Q.^  | 

etc.        I 

Rem.  The  square  root  of  2  may  be  employed  in  finding  a 
right  angle,  because  a  right  angle  (by  Geometry)  can  be  formed 
by  three  pieces  of  lengths  proportional  to  1,  1,  ^2.  If  we 
make  the  lengths  12,  12,  17,  the  error  will,  by  Cor.  V,  be  less 

than  zt^wt:,  or  less  than  -—  of  the  whole  length. 

EXERCISES. 

Develop  the  following  square  roots  as  continued  fractions, 
and  find  six  or  more  of  the  partial  fractions  approximating  to 
each : 

I.     V3.  2.     ^5.  3.     \/6.  4.     VlO. 

5.  Develop  a  root  of  the  quadratic  equation 
x^  —  ax  —  l=^0, 
commencing  the  operation  by  dividing  the  equation  by  x. 

Periodic  Continued  Fractions. 

349.  Def,  A  Periodic  continned  fraction  is  one  in 
which  the  denominators  repeat  themselves  in  regular 
order. 


CONTINUED   FRACTIONS,  271 

Example.     A  continued  fraction  in  which  the  successive 
denominators  are 

2,    3,    5,    2,    3,    5,    2,    3,    5,    etc.,    ad  infinitum, 
is  periodic. 

A  periodic  continued  fraction  can  be  expressed  as 
the  root  of  a  quadratic  equation. 


I. 


EXAMPLES. 
1 


i  +  i 


2  +  ^ 


1  +  ^ 


2  +  etc. 

If  we  put  X  for  the  value  of  this  fraction,  we  have 
1 


X  = 


1+' 


2  +  a; 
We  find  the  value  thus: 

1,        2-{-x. 

0  1  2  +  a; 

1'  1'  Z^-x 

Because  this  expression  is  x  itself,  we  have 

^  -^  3  +  0?^ 
which  reduces  to  the  quadratic  equation 

a;3  +  2a;  —  2. 

2.  Let  us  take  the  fraction  of  which  the  successive  denom- 
inators are  2,  3,  5,  2,  3,  5,  etc.,  namely, 

1 


X  = 


a+i 


s  +  l 


^     1 

5  +  - 


2  +  ^ 


3  +  etc.^ 


272  GONTINUED   FRACTIONS, 


or. 


JU   — 

.+'  , 

^  +  5  +  .- 

We  compute  thus: 

2,             3, 

a;  +  5. 

0              1 

3              dx  +  lQ 

1'            2' 

1'            72; +  37' 

Hence  we  have,  to  determine  x,  the  quadratic  equation. 

Zx  +  16 

or        W  +  34:t^  =  16. 

7a;  +  37 

350.  Development  of  the  Root  of  a  Quadratic  Eq^iation, 
A  root  of  a  quadratic  equation  may  be  developed  in  a  continued 
fraction  by  the  following  process.  Let  the  equation  in  its 
normal  form  be  (§  192), 

mx^  +  nx  +  p  =1  0,  (1) 

m,  n,  and  p  being  whole  numbers.     We  shall  then  have 

_  —  n  ±:  Vn^  —■  ^mp 
'  "^  2m 

Let  a  be  the  greatest  whole  number  in  x,  which  we  may 
find  either  by  trial  in  (1)  or  by  this  value  of  x.    Then  assume 

1 

X  =  a  -\ • 

^1 

and  substitute  this  value  of  x  in  the  original  equation.  Then, 
regarding  x^  as  the  unknown  quantity,  we  reduce  to  the  nor- 
mal form,  which  gives 

{ma^  +  na  +p)x^'^  +  {2ma  -{-  n)xi  +  m  =  0.         (2) 
It  a^  is  the  greatest  whole  number  in  x^,  we  put 

and  by  substituting  this  value  of  x^  in  (2),  we  form  an  equa- 
tion in  x^.  Continuing  the  transformations,  we  find  the 
greatest  whole  number  in  x^,  which  will  be  called  a^^  and  so  on. 
The  root  will  then  be  expressed  as  a  whole  number  a  plus 
the  continued  fraction  of  which  the  denominators  sltg  a^,  a^, 
a^,  etc. 


BOOK    X. 
THE     COMBINATORY    ANALYSIS. 


CHAPTER     I. 

PE  R  M  U  TATI  0  N  S. 


351.  Def.  The  different  orders  in  which  a  number 
of  things  can  be  arranged  are  called  their  Permuta- 
tions. 

Examples.    The  permutations  of  the  letters  a,  h,  are 

ab,     ha. 
The  permutations  of  the  numbers  1,  2,  and  3  arc 
123,     132,     213,     231,     312,     321. 

Problem.  To  find  how  many  pei^mutations  of  any 
given  nu7)^her  of  things  are  possible. 

Let  us  put 

Fi,  the  number  of  permutations  of  i  things. 

It  is  evident  from  the  first  of  the  above  examples  that  there 
are  two  permutations  of  two  things.    Hence, 

To  find  the  permutations  of  three  letters,  a,  b,  c,'vie  form 
three  sets  of  permutations,  the  first  beginning  with  a,  the  sec- 
ond with  b,  and  the  third  with  c. 

In  each  set  the  first  letter  is  to  be  followed  by  all  possible 
permutations  of  the  remaining  letters,  namely : 

In   1st  set,  after  a  write  be,  cb,  making  abc,  acb. 

"    2d     "        "  b  "  ac,  ca,         "  bac,  bca. 

"    3d      "        "  c  "  ab,  ba,         "  cab,  cba. 
18 


274  PERMUTATIONS. 

Hence,  Pg  =  3-2  =  6. 

The  permutations  of  n  things  can  be  divided  into  sets. 
The  first  set  begins  with  the  first  thing,  followed  by  all  possi- 
ble permutations  of  the  remaining  n  —  1  things,  of  which  the 
number  is  Pn-i*  The  second  set  begins  with  the  second  thing, 
followed  by  all  possible  permutations  of  the  remaining  n  —  1 
things,  of  which  the  number  is  also  Pn-i,  and  so  with  all  n 
sets.  Hence,  whatever  be  n,  there  will  be  n  sets  of  Pn-i  per- 
mutations in  each  set.     Therefore, 

Pfi  =  nPn~i. 
This  equation  enables  us  to  find  P^  whenever  we  know 
Pn-h  and  thus  to  form  all  possible  values  of  Pn,  as  follows : 


It  is  evident  that 

P,  =1. 

We  have  found 

Pg  =:  2.1  =  2! 

((                         6C 

Pg   =  3.2.1  =.  3!  =::  6. 

Putting  ^  =  4,  we  have 

P^  =:  4P3  =  4!  ==  24. 

71  =  5,    ''     '' 

P5    =r  5P4    r=   5  !  =   120. 

etc. 

etc.                    etc. 

It  is  evident  that  the  number  of  permutations  of  n  things 
is  equal  to  the  continued  product 

1.2.3.4 n, 

which  we  have  represented  by  the  symbol  n !  so  that 

Pn  =  n\ 

EXERCISES.* 

1.  Write  all  the  permutations  of  the  following  letters  : 

bed,        acd,        abd^        abed. 

2.  What  proportion  of  the  possible  permutations  of  the 
letters  a,  e,  m,  t,  make  well-known  English  words  ? 

3.  Write  all  the  numbers  of  four  digits  each  of  which  can 
be  formed  by  permuting  the  four  digits  1,  2,  3,  4. 

4.  How  many  numbers  is  it  possible  to  form  by  permuting 
the  six  figures  1,  2,  3,  4,  5,  6. 

*  If  the  student  finds  any  difficulty  in  reasoning  out  these  exercises, 
he  is  recommended  to  try  similar  cases  in  which  few  symbols  are  involved 
by  actually  forming  the  permutations,  until  he  clearly  sees  the  general 
principles  involved. 


PERMUTATIONS.  275 

5.  At  a  dinner  party  a  row  of  6  plates  is  set  for  the  host 
and  5  guests.  In  how  many  ways  may  they  be  seated,  subject 
to  the  condition  that  the  host  must  have  Mr.  Brown  on  his 
right  and  Mr.  Hamilton  on  his  left  ? 

6.  Of  all  numbers  that  can  be  formed  by  permuting  the 
seven  digits,  1,  2  ....  7 : 

{a)  How  many  will  be  even  and  how  many  odd  ? 
(l)  In  how  many  will  the  seven  digits  be  alternately  even 
and  odd? 

{c)  In  how  many  will  the  three  even  digits  all  be  together  ? 
{d)  In  how  many  will  the  four  odd  digits  all  be  together  ? 

7.  In  how  many  permutations  of  the  8  letters,  a,  i,  c,  d,  e, 
/,  g,  h,  will  the  letters  d,  e,  /,  stand  together  in  alphabetical 
order  ? 

8.  In  how  many  of  the  above  permutations  will  the  word 
deaf  be  found  ? 

9.  In  how  many  of  the  permutations  of  the  first  9  letters 
will  the  words  age  and  hid  be  both  found  ? 

10.  A  party  of  5  gentlemen  and  5  ladies  agree  with  a  math- 
ematician to  dance  a  set  for  every  way  in  which  he  can  divide 
them  into  couples.     How  many  sets  can  he  make  them  dance  ? 

11.  In  how  many  of  the  permutations  of  the  letters  a,  h,  c, 
d,  e,  will  d  and  no  other  letter  be  found  between  a  and  e, 

12.  In  how  many  of  the  permutations  of  the  six  symbols, 
a,  h,  Cy  d,  e,f,  will  the  letters  abc  be  found  together  in  one 
group,  and  the  letters  def  in  another  ? 

13.  How  many  permutations  of  the  seven  symbols,  a,  h,  c, 
d,  e,  /,  g,  are  possible,  subject  to  the  condition  that  some  per- 
mutation of  the  letters  abc  must  come  first  ? 

14.  The  same  seven  syjnbols  being  taken,  how  many  per- 
mutations can  be  formed  in  which  the  letters  abc  shall  remain 
together  ? 

Permutations  of  Sets. 

352.  Def,  When  permutations  are  formed  of  only 
s  things  out  of  a  whole  number  n,  they  are  called  Per- 
mutations of  n  things  taken  ^  at  a  time. 


276  PERMUTATIONS. 

Example.  The  permutations  of  the  three  letters  a,  h^  c, 
taken  two  at  a  time,  are 

ab,    bUy    aCy     ca,    Ic,     cb. 

The  peimutations  of  1,  2,  3,  4,  taken  two  at  a  time,  are 

12,  13,  14,  21,  23,  24,  31,  32,  34,  41,  42,  43. 

Pkoblem.  To  find  the  number  of  permutations  of 
n  things  taken  s  at  a  time. 

Suppose,  first,  that  we  take  two  things  at  a  time,  as  in  the 
above  examples.  We  may  choose  any  one  of  the  n  things  as 
the  first  in  order.  Which  one  soever  we  take,  we  shall  have 
n  —  1  left,  any  one  of  which  may  be  taken  as  the  second  in 
order.     Hence,  the  permutations  taken  2  at  a  time  will  be 

n{n  —  l). 

[Compare  with  the  last  example,  where  7i  =  4.] 
To  form  the  permutations  3  at  a  time,  we  add  to  each  per- 
mutation by  2's  any  one  of  the  n  —  2  things  which  are  left. 
Hence,  the  number  of  permutations  3  things  at  a  time  is 

n{n  —  l){n  —  2). 

In  general,  the^  permutations  of  7i  things  taken  5  at  a  time 
will  be  equal  to  the  continued  product  of  the  s  factors, 

n  {n  —  1)  (^  —  2) {n  —  s  -\-l), 

n\ 

which  is  equal  to  the  quotient    -, 

(n — s)\ 

It  will  be  remarked  that  when  5  =  /z,  we  shall  have  the 
case  already  considered  of  the  permutations  of  all  n  things. 

EXE  RCI  SES. 

1.  Write  all  the  numbers  of  two  figures  each  which  can  be 
formed  from  the  four  digits,  3,  5,  7,  9. 

2.  Write  all  the  numbers  of  three  figures,  beginning  with 
1,  which  can  be  formed  from  the  five  digits,  1,  2,  3,  4,  5. 

3.  How  many  different  numbers  of  four  figures  each  can 
be  formed  with  the  digits  1,  2,  3,  4,  5,  6,  no  figure  being  re- 
peated in  any  number  ? 


PERMUTATIONS. 


277 


4.  Explain  how  all  the  numbers  in  the  preceding  exercise 
may  be  written,  showing  how  many  numbers  begin  with  1, 
how  many  with  2,  etc. 

5.  In  how  many  ways  can  3  gentlemen  select  their  partners 
from  5  ladies  ? 

6.  How  many  even  numbers  of  3  different  digits  each  can 
be  formed  from  the  seven  digits,  1,  2,  ....  7  ? 

7.  How  many  of  these  numbers  will  consist  of  an  odd 
digit  between  two  even  ones  ? 


Circular  Permutations. 

353.  If  we  have  the  three  letters  a,  d,  c,  arranged  in  a 
circle,  as  in  the  adjoining  figure,  then, 
however  we  arrange  them,  we  shall  find 
them  in  alphabetical  order  by  beginning 
with  a  and  reading  them  in  the  suitable 
direction.  Hence,  there  are  only  two 
different  circular  arrangements  of  three 
letters  instead  of  six,  namely,  the  two 
directions  in  which  they  may  be  in  al- 
phabetical order. 

Next  suppose  any  number  of  symbols,  say  a,  I,  c^d,  e,f,  g^ 
h,  and  let  there  be  an  equal  number  of  positions  around  the 
circle  in  which  they  may  be  placed.  These  positions  are  num- 
bered 1,  2,  3,  4,  5,  6,  7,  8. 

For  every  arrangement  of  the  sym- 
bols we  may  turn  them  round  in  a  body 
without  changing  the  arrangement. 
Each  symbol  will  then  pass  through  all 
eight  positions  in  succession. 

By  performing  this  operation  with 
every  arrangement,  we  shall  have,  all 
possible  permutations  of  the  eight  things 
among  the  eight  positions,  the  number 

of  which  is  8 ! ,  which  are  therefore  eight  times  as  many  as 
the  circular  arrangements. 


278  PERMUTATIONS. 

Hence  the  number  of  different  circular  arrangements  is 

8? 

— ' ,  which  is  the  same  as  7 ! 
o 

In  general,  if  we  represent  the  number  of  circular  arrange- 
ments of  n  things  by  Gn,  we  shall  have 

Gn  =  {n-1)\ 

The  same  result  maybe  reached  by  the  following  reasoning. 
To  form  a  circular  arrangement,  we  may  take  any  one  symbol, 
a  for  example,  put  it  into  a  fixed  position,  say  (1),  and  leave  it 
there. 

All  possible  arrangements  of  the  symbols  will  then  be 
formed  by  permuting  the  remaining  symbols  among  the  re- 
maining positions.     Hence, 

On  =  Fn-1  =  {n  -  1) ! 
as  before. 

EXERCISES. 

1.  In  how  many  orders  can  a  party  of  7  persons  take  their 
places  at  a  round  table? 

2.  In  how  many  orders  can  a  host  and  7  guests  sit  at  a 
round  table  in  order  that  the  host  may  have  the  guest  of  high- 
est rank  upon  his  right  and  the  next  in  rank  on  his  left? 

3.  Five  works  of  four  volumes  each  are  to  be  arranged  on 
a  circular. shelf.  How  many  arrangements  are  possible  which 
will  keep  the  volumes  of  each  set  together  and  in  proper  order, 
it  being  indifferent  in  which  direction  the  numbers  of  the 
volumes  read. 

4.  In  how  many  circular  arrangements  of  the  5  letters  a,  d, 
c,  d,  e,  will  a  stand  between  h  and  d  ? 

5.  If  the  10  digits  are  to  be  arranged  in  a  circle,  in  how 
many  ways  can  it  be  done,  subject  to  the  condition  that  even 
and  odd  digits  must  alternate  ?     (Note  that  0  is  even.) 

6.  The  same  thing  being  supposed,  how  many  arrange- 
ments are  possible,  subject  to  the  condition  that  the  even  digits 
must  be  all  together  ? 

7.  In  how  many  circular  arrangements  of  the  first  six  let- 
ters will  the  word  deaf  he  found?  What  will  be  the  difference 
of  the  results  if  you  are  allowed  to  spell  it  in  either  direction? 


PERMUTATIONS.  279 

Permutations  when  Several  of  the  Thing^s  are 
Identical. 

254.  If  the  same  thing  appears  several  times  among  the 
things  to  be  permuted,  the  number  of  different  permutations 
will  be  less  than  when  the  things  are  all  different. 

Example.     The  permutations  of  aahh  are 

aahl,     abab,     ahha,    baab,     haba,    bbaa,  (1) 

which  are  only  six  in  number. 

Problem.  To  firul  the  number  of  permutations  when 
several  of  the  things  are  identical. 

Let  us  first  examine  how  all  24  permutations  of  4  things 
may  be  formed  from  the  above  6  permutations  of  aabb.  Let 
us  distinguish  the  two  a's  and  the  two  ^'s  by  accenting  one  of 
each.  Then,  from  each  permutation  as  written,  four  may  be 
formed  by  permuting  the  similar  letters  among  themselves. 
For  example,  taking  abba,  and  writing  it  abb' a',  we  have,  by 
permuting  the  similar  letters, 

ab'ba!,     a'b'ba.     abb'a',     a'bVa,  (2) 

In  the  same  way  four  permutations,  differing  only  in  the 
arrangement  of  the  accents,  may  be  formed  from  each  of  the 
6  permutations  (1),  making  24  in  all,  as  there  ought  to  be. 
(§  251.) 

Generalizing  the  preceding  operation,  we  reach  the  follow- 
ing solution  of  our  problem.  Let  the  symbols  to  be  permuted 
be  a,  b,  c,  etc. 

Suppose  that  a  is  repeated  r  times, 

etc.  etc.  etc. 

and  let  the  whole  number  of  symbols,  counting  repetitions,  be 
n,  so  that 

n  =  r'j-s-\-t-\-  etc. 

[In  the  preceding  example  (1),  tz  —  4,  r  =  2,  .9  =  2.] 
Also  put  Xn,  the  required  number  of  different  permutations 
of  the  71  symbols. 


280  PERMUTATIONS. 

Suppose  these  Xn  different  permutations  all  written  out, 
and  suppose  the  symbols  which  are  repeated  to  be^distinguished 
by  accents.     Then: 

From  each  of  the  Xn  permutations  may  be  formed  Pr=^r\ 
permutations  by  permuting  the  «'s  among  themselves,  as  in 
(2).     We  shall  then  have  r !  Xn  permutations. 

From  each  of  the  latter  may  be  formed  s\  permutations  by 
permuting  the  Z>'s  among  themselves.  We  shall  then  have 
5 !  r !  X  Xn  permutations. 

From  each  of  these  may  be  found  t !  permutations  by  in- 
terchanging the  d^  among  themselves. 

Proceeding  in  the  same  way,  we  shall  have 

Xn  X  r\  X  s\  X  t\  X  etc. 
possible  permutations  of  all  7i  things.     But  this  number  has 
been  shown  to  be  n\     Therefore, 

Xn  X  r\  X  s\  X  t\  X  etc.  =  n ! 

By  division,  X„  =  ^^j^^----,  (3) 

which  is  the  required  expression. 

We  remark  that  if  any  symbols  are  not  repeated,  the  for- 
mula (3)  will  still  be  true  by  supposing  the  number  correspond- 
ing to  r,  s,  or  t  to  be  1. 

EXAMPLES. 

I.  The  number  of  possible  permutations  of  aabl  are 
z=z  — —  =r  6,  as  already  found. 


2!  2!  ~  2-2 
2.  The  possible  permutations  of  aaahhcd  are 
7 !  5040 


3!  2!         6-2 


420. 


EXERCI  SES. 

Write  all  the  permutations  of  the  letters: 
I.     aaah,  2.     aabc,  3.     aaabc. 

4.  How  many  different  numbers  of  seven  digits  each  can 
be  formed  by  permuting  the  figures  1112225  ? 


PERMUTATIONS,  281 

5.  If  every  different  permutation  of  letters  made  a  word, 
how  many  words  of  13  letters  each  could  be  formed  from  the 
word  Massachusetts. 

The  Two  Classes  of  Permutations. 

255.  The  n\  possible  permutations  of  n  things  are  divisi 
ble  into  two  classes,  commonly  distinguished  as  even  permu- 
tations and  odd  permutations  in  the  following  way: 

We  suppose  the  7i  things  first  arranged  in  alphabetical  or 
numerical  order, 

a,d,Cyd,,,.,        or        1,  2,  3,  4,  ...  .  n, 

and  we  call  this  arrangement  an  evmi  permutation. 

Then,  having  any  other  permutation,  we  count  for  each 

thing  how  many  other  things  of  lower  order  come  after  it,  and 

take  the  sum. 

If  this  sum  is  even,  the  permutation  is  an  even  one ;  if  odd, 

an  odd  one. 

EXAMPLES. 

1.  Consider  the  permutation  265143. 

Here  2  is  followed  by  1  number  of  lower  order,  namely,  1. 
"     Q     "         "         4       "         "  "  "       5,1,4,3. 

"    5     "        "         3       "        "         ''  "       1,4,3. 

a       \       "  ^*  Q  ''  '*•  ^< 

a      4      ^^  "  1         ''  ''  '^  ^^         3 

Then  1+4  +  3  +  0  +  1  =  9.     Hence  the  permutation  is  odd. 

2.  Consider  cdbea. 

Here  c  is  followed  by  2  letters  before  it  in  order,  namely,  ba, 
''    d  ,"  "  2       "  "  "  "        la. 

6i         Jj  ii  6i  1  (<  66  66  66  ^ 

66  Q  "  ''  1  ^^  ^^  ^^  ^^  n 

Then  2  +  2  +  1  +  1  =  6.     Hence  the  permutation  is  even. 

Def,  The  total  number  of  times  which  a  thing  less 
in  order  follows  one  greater  in  order  is  called  the 
Number  of  Inversions  in  a  permutation. 


282  PERMUTATIONS. 

Example.  In  the  preceding  permutation^  265143,  the 
number  of  inversions  is  9.     In  cdhed  it  is  6. 

Eem.  It  will  be  seen  that  the  class  of  a  permutation  is 
even  or  odd,  according  as  the  number  of  inversions  is  even  or 
odd. 

Theorem  I.  If,  in  a  perinutaUon,  two  things  are 
interchanged,  the  class  will  he  changed  from  even  to  odd, 
or  from  odd  to  even. 

Proof.  Consider  first  the  case  in  which  a  pair  of  adjoining 
things  are  interchanged.     Let  us  call : 

ilc,  the  two  things  interchanged. 

A,  the  collection  of  things  which  precede  i  and  Jc. 

C,  the  collection  of  things  which  follow  them. 

The  first  permutation  will  then  be 

AikC.'^  {a) 

After  interchanging  i  and  k,  ifc  will  be 

AUG.  \b) 

Because  the  order  of  things  in  A  remains  undisturbed,  each 
thing  in  A  is  followed  by  the  same  things  as  before.  In  the 
same  way,  each  thing  in  G  is  preceded  by  the  same  things  as 
before. 

Hence,  the  number  of  times  that  each  thing  in  A  or  G  is 
followed  by  a  thing  less  in  order  remains  unchanged,  and, 
leaving  out  the  pair  of  things,  i,  k,  the  number  of  inversions 
is  unchanged. 

But,  by  interchanging  i  and  k,  the  new  inversion  ki  is  in- 
troduced. Therefore  the  number  of  inversions  is  increased 
by  1. 

*  This  form  of  algebraic  notation  differs  from  those  already  used  in 
that  the  symbols  A  and  C  do  not  stand  for  quantities,  but  mere  collec- 
tions of  letters.  It  is  an  application  of  the  general  principle  that  a  single 
symbol  may  be  used  to  represent  any  set  of  symbols,  but  must  represent 
the  same  set  throughout  the  same  question.  A  and  C  are  here  used  to 
show  to  the  eye  that  in  forming  the  permutations  of  (5)  from  {a),  all  the 
letters  on  each  side  of  ik  preserve  their  relative  positions  unchanged. 


PERMUTATIONS.  283 

If  the  first  arrangement  is  Id,  this  one  inversion  is  removed. 
Hence^  in  either  case  the  number  of  inversions  is  changed  by 
1,  and  is  therefore  changed  from  odd  to  even,  or  vice  versa. 

Illustration,     In  the  permutation 
265143, 
the  inversions,  as  already  found,  are  the  following  nine  : 
21,     65,     61,     64,     63,     51,     54,     53,     43. 
Let  us  now  interchange  5  and  1,  making  the  permutation 

261543. 
The  inversions  now  are  ^ 

21,     61,     65,     64,     63,     54,     53,     43, 
the  same  as  before,  except  that  51  has  been  removed. 

Next  consider  the  case  in  which  the  things  interchanged 
do  not  adjoin  each  other.     Suppose  that  in  the  permutation 

h  a  d  e  h  c  f 

we  interchange  a  and  h.  We  may  do  this  by  successively  in- 
terchanging a  with  d,  with  e,  and  with  h,  making  three  inter- 
changes, producing 

I  d  e  h  a  c  f . 

Then  we  interchange  h  with  e  ^and  with  d,  making  two 
interchanges,  and  producing 

h  h  d  e  a  c  f , 

which  effects  the  required  interchange  of  a  with  h. 

The  number  of  the  neighboring  interchanges  is  3-|-2  r=  5, 
an  odd  number.  Because  the  number  of  inversions  is  changed 
from  odd  to  even  this  same  odd  number  of  times,  it  will  end 
in  the  opposite  class  with  which  it  commenced. 

Theorem  II.  I7^e  possible  permutations  of  n  things 
are  one-half  even  and  one-half  odd. 

Proof,    Write  the  n !  possible  permutations  of  the  n  things. 

Then  interchange  some  one  pair  of  things  {e.g.,  the  first 
two  things)  in  each  permutation.  We  shall  have  the  same 
permutations  as  before,  only  differen  ]y  arranged. 


1  3 

odd. 

1  -i 

even. 

2  3 

even. 

3  1 

odd. 

3  2 

odd. 

3  1 

even. 

284  PERMUTATIONS. 

By  the  change,  every  even  permutation  will  be  changed  to 
odd,  and  every  odd  one  to  even. 

Because  every  odd  one  thus  corresponds  to  an  even  one, 
and  vice  versa,  their  numbers  must  be  equal. 

Illustration.  The  permutations  in  the  second  column  fol- 
lowing  are  formed  from  those  in  the  first  by  interchanging  the 
first  two  figures  : 

12  3        even,  2 

13  2         odd,  3 
2  13         odd,  1 

2  3  1        even,  3 

3  12        even,  1 
3  2  1        odd,  2 

EXERCISES. 

Count  the  number  of  inversions  in  each  of  the  following 
permutations ; 

I.     hcdagef.  2.     Magdef.  3.     325941. 

4.     5432.  5.     82917364.  6.     82971364. 

S56  Def.  A  Symmetric  Function  is  one  which  is 
not  changed  by  permuting  the  symbols  which  enter  into  it. 

An  Alternating  Fuhction  is  one  which,  when  any  two 
of  its  symbols  are  interchanged,  changes  its  sign  without 
changing  its  absolute  value. 

EXERCISES 

Show  which  of  the  following  functions  are  symmetric  and 
which  are  alternating : 

I,  a  +  6  +  c.  2.     ahc. 

3.  ^  (6  +  c)  -f  5  (c  +  a)  +  c  (a  +  6), 

4.  a^  (b  -  c)  +  h'  (c-a)  +  c'  {a  -  b). 

5.  a'  {b  +  c)  +  b'  (c+a)  -he'  (a  +  5). 

6.  (a-b)  {b  -  c)  (c  -  a), 

7.  ab  -\-  be  -{-  oa. 


COMBINATIONS.  285 

CHAPTER     II. 

COMBINATIONS. 

357.  Def,  The  number  of  ways  in  which  it  is  pos- 
sible to  select  a  set  of  .^  things  out  of  a  collection  of  n 
things  is  called  the  Number  of  Combinations  of  s 
things  in  n. 

Ex.  I.  From  the  three  symbols  a,  h,  c,  may  be  formed  the 

couplets, 

ah,        acy         he. 

Hence  there  are  three  combinations  of  2  things  in  3. 

Ex.  2.  From  a  stud  of  four  horses  may  be  formed  six  dif- 
ferent span.    If  we  call  the  horses  A,  B,  C,  J),  the  dijBFerent 

span  will  be 

AB,     AC,     AD,     BC,     BD,     CD. 

Rem.  1.  A  set  is  regarded  as  different  when  any  one  of  its 
separate  things  is  different. 

Eem.  2.  Combinations  differ  from  permutations  in  that, 
in  forming  a  combination,  no  account  is  taken  of  the  order  of 
arrangement  of  things  in  a  set.  For  instance,  ah  and  ha  are 
the  same  combination.  Hence,  we  may  always  suppose  the 
letters  or  numbers  of  a  combination  to  be  written  in  alpha- 
betical or  numerical  order. 

Notation,  The  number  of  combinations  of  s  things  in  7i 
is  sometimes  designated  by  the  symbol, 

c» 

Problem.  To  find  the  ninnher  of  combinations  of  s 
things  in  n. 

If  we  form  every  possible  set  of  s  things  out  of  n  things, 
and  then  permute  the  s  things  of  each  set  in  every  possible 
way,  we  shall  have  all  the  permutations  of  n  things  taken  s  at 
a  time  (§  252).     That  is, 

Cg     X   Is 


286  GOMBmATIONS. 

express  the  number  of  permutations  of  n  things  taken  5  at  a 
time.     But  we  have  found  this  number  to  be 


n  (n  —  1)  (m  —  2)  .  .  .  .  (m  —  s  +  1). 

We  have  also  found 

P,  =  .s!  =  1.2-3-4 s. 

Hence,     Cf  xsl  =  n{n  —  l)  {iu—  2)  ....  (re  - 

-  s  +  1), 

,                           n  _  M  («  -  1)  (w  -  2) {n  - 

-s  +  1) 

^'   -                   1.2.3. 4....  s 

=  g)  (§  228,  3) ; 

C^=     .   ,^'      ... 

which  is  the  required  expression. 

Rem.  For  every  combination  of  s  things  which  we 
can  take  from  n  things,  a  combination  ot  n—  s  things 
will  be  left. 

Hence,  Cf  =  Cls. 

This  formulae  may  be  readily  derived  from  the  expression 
for  the  number  of  combinations.     For,  if  we  take  the  equation 

pn  _  ^ 

^^    -  s\  {n-s)V 

this  formula  remains  unaltered  when  we  substitute  n  —  s  for 
s,  and  therefore  also  represents  the  combinations  of  7i  —  s 
things  in  n. 

Def.  Two  combinations  which  together  contain  all 
the  things  to  be  combined  are  called  two  Complement- 
ary combinations. 

EXERCISES. 

1.  Write  all  combinations  of  two  symbols  in  the  five  sym- 
bols, a^  h,  Cy  d,  e. 

2.  Write  all  combinations  of  three  symbols  in  the  same 
letters,  and  show  why  the  number  is  the  same  as  in  Ex.  i. 


GOMBINxiTIONS.  287 

3.  A  span  of  horses  being  different  when  either  horse  is 
changed,  how  many  different  span  may  be  formed  from  a  stud 
of  3  ?     Of  7  ?     Of  9  ? 

4.  If  four  points  are  marked  on  a  piece  of  paper,  how  many 
distinct  lines  can  be  formed  by  joining  them,  two  and  two  ? 
How  many  in  the  case  of  n  points  ? 

From  each  one  of  the  points  can  be  drawn  7i  -—1  lines  to 
other  points;  then  why  are  there  not  71  {n  —  1)  Hnes? 

5.  If  five  lines,  no  two  of  which  are  parallel,  intersect  each 
other,  how  many  points  of  intersection  will  there  be  ?  How^ 
many  in  the  case  of  n  lines  ? 

6.  If  n  straight  lines  all  intersect  each  other,  how  many 
different  triangles  can  be  found  in  the  figure  ? 

7.  In  how  many  different  ways  may  a  set  of  four  things  be 
divided  into  two  pairs  ? 

8.  In  how  many  ways  can  a  party  of  four  form  partners  at 
whist? 

9.  In  how  many  ways  can  the  following  numbers  be  thrown 
with  three  dice : 

{a)     1,1,1;  {d)     1,2,2;  {c)     1,2,3. 

10.  A  school  of  15  young  ladies  have  the  privilege  of  send- 
ing a  party  of  5  every  day  to  a  picture  gallery,  provided  they 
do  not  send  the  same  party  twice.  How  many  visits  can  they 
make? 

Combinations  with  Repetition. 

358.  Sometimes  combinations  are  formed  with  the  liberty 
to  repeat  the  same  symbol  as  often  as  we  please  in  any  set. 

Example.  From  the  three  things  a,  h,  c,  are  formed  the 
six  combinations  of  two  things  with  repetition, 

aa,        ah,        ac,        hh,         he,         cc, 

Pkoblem.  To  -find  the  ninnhcr  of  comhinations  of  s 
things  in  n,  wheiz  repetition  is  allowed. 

Solution.     Let  the  n  things  be  the  first  71  numbers, 
1,  2,  3,  4,  ...  .  u. 


288  COMBINATIONS. 

Form  all  possible  sets  of  s  of  these  numbers  with  repetition, 
the  numbers  of  each  set  being  arranged  in  numerical  order. 

Let  Rs  be  the  required  number  of  sets.     Then,  in  each  set, 

Let  the  first  number  stand  unchanged. 
Increase  the  2d  number  by  1. 

"         "    3d       "         "  2. 

''        ''    4th     "         "  3. 


We  shall  then  have  Bs  sets  of  s  numbers,  each  without  rep- 
etition. 

Example.    From  the  numbers  1,  2,  3  are  formed  with  repetition, 

11,  12,    13,    22,    23,    33. 
Then,  increasing  the  second  numbers  by  1,  we  have 

12,  13,    14,    23,    24,    34. 

The  greatest  possible  number  in  any  set  after  tlie  increase 
will  be  n  -{-  s  —  1,  because  the  greatest  number  from  which 
the  selection  is  made  is  n,  and  the  greatest  quantity  added  is 
5—1.  Hence  all  the  new  sets  will  consist  of  combinations  of 
s  numbers  each  from  the  n  -\-  s  —  1  numbers, 

1,  2,  3,  4, ....  :?^ ....  :^  +  5  —  1.  {a) 

No  two  of  these  combinations  can  be  the  same,  because  then 
two  of  the  original  combinations  would  have  to  be  the  same. 
Hence  the  new  sets  are  all  different  combinations  of  s  numbers 
from  the  n  -[-  s  —  1  numbers  {a).  Therefore  the  number  of 
combinations  cannot  exceed  the  quantity  (7/^. 

Conversely,  if  we  take  all  possible  combinations  of  s  differ- 
ent numbers  in  n  -{-  s  —  1,  arrange  each  in  numerical  order, 
and  subtract  1  from  the  second,  2  from  the  third,  etc.,  we 
shall  have  different  combinations  from  the  first  n  numbers 
with  repetitions.  Hence  the  number  of  combinations  in  the 
second  class  cannot  exceed  those  of  the  first  class. 

Hence  we  conclude  that  the  number  of  combinations  of  s 
things  in  n  with  repetition  is  the  same  as  the  combinations  of 
s  things  in  n  -{-  s  —  1  without  repetition,  or 


COMBINATIONS.  289 


»7i j-iTh^s-i  In  -{-  s  —  1\ 


__  n{n  +  l){n  -\-2) {n-\-  s  —  1) 

""  1.2.3.4 s 

EXERCISES. 

1.  Write  all  possible  combinations  of  3  numbers  with  repe- 
tition out  of  the  three  numbers  1, 2, 3 ;  then  increase  the  second 
of  each  combination  by  1  and  the  third  by  2,  and  show  that 
we  haye  all  the  combinations  of  three  different  numbers  out  of 
1,2,3,4,5. 

2.  How  many  combinations  of  4  things  in  4  with  repeti- 
tion ?     Oi  n  things  in  tz  ? 

In  the  last  question  and  in  the  following,  reduce  the  result  to  its 
lowest  terms. 

3.  How  many  combinations  of  7^  + 1  things  m.  n—l  with 
repetition  ? 

Special  Cases  of  Combinations. 

359.  It  is  plain  that 

because  each  of  these  combinations  consist  simply  of  one  of  the 
n  things.     Hence,  also, 

Cl-i  =  n, 

because  in  every  such  combination  one  letter  is  omitted. 
It  is  also  plain  that 

because  the  only  combination  of  7i  letters  is  that  comprising 
the  n  letters  themselyes.    Hence  we  write,  by  analogy, 

G^  =  1, 

although  a  combination  of  nothing  does  not  fall  within  the 
original  definition  of  a  combination. 

360.  The  formulae  of  combinations  sometimes  enable  us 
to  discover  curious  relations  of  numbers. 

1.  Let  us  inquire  how  we  may  form  the  combinations  of 


*290  GOMBINAriONS. 

s  -\-  1  things  when  we  have  those  of  5  things.  Let  the  n 
things  from  which  the  combinations  are  to  be  formed  be  the 
letters 

a,  d,  c,  d,  e,f,  g,  etc {n\n  number). 

Let  all  the  combinations  of  5  + 1  of  these  n  letters  be  writ- 
ten in  alphabetical  order.     Then : 

1.  In  the  combinations  beginning  with  a,  the  letter  a  will 
be  followed  by  all  possible  combinations  of  s  letters  out  of  the 
71  —  1  letters  d,  c,  d,  etc.,  of  which  the  number  is  Cf'^. 

2.  In  the  combinations  beginning  with  h,  the  letter  h  is 
followed  by  all  combinations  of  s  letters  out  of  the  n  —  2  let- 
ters c,  d,  e,  f,  etc.  Therefore  there  are  G'^'^  combinations 
beginning  with  h, 

3.  In  the  same  way  it  may  be  shown  that  there  are  C'^~^ 
combinations  beginning  with  c,  C^~^  beginning  with  d,  etc. 
The  series  will  terminate  with  a  single  combination  of  the  last 
s  +  1  letters. 

Since  we  thus  have  all  combinations  of  s  +  1  letters,  we 
find,  by  summing  up  those  beginning  with  the  several  letters 
a,  b,  c,  etc., 

or'  +  cr'  +  or'  +  ....  +  ci  =  c»+i.      («) 

Substituting  for  the  combinations  their  values,  we  find 

By  the  notation  (§  228,  3),  all  the  terms  of  the  first  member 
have  the  common  denominator  s ! ,  while  the  numerators  are 
each  composed  of  the  factors  of  s  consecutive  numbers.  Mul- 
tiplying both  sides  by  s !  and  reversing  the  order  of  terms  in 
the  first  member,  we  have 

1.2.3 5  +  2-3.4 5  +  1  +  etc. 

etc.  etc. 

+  (n  —  s  —  l) (^  _  3)  (^  —  2) 

+  {n  —  s) {71  —  2)  {71  —  1) 

_  {n  —  s)  .  ,  ,  ,  {n  —  2)  (n  —  1)  ^ 

_  ____ 


\ 


COMBINATIONS.  291 

The  student  is  now  recommended  to  go  over  the  preceding  process 
with  special  simple  numerical  values  of  n  and  s  which  he  may  select  for 
himself. 

EXAMPLES. 

If  ^  =  5  and  s  =  2,  we  have 

1.2  +  2.3  +  3.4  =  ^. 

6 

If  II  =  7  and  5  =  3. 

1.2.3  +  2.3.4  -f  3.4.5  -f-  4.5.6  =  i:^!^-. 

If  iz  =2  7  and  s  =  4, 

1.2.3.4  +  2.3.4.5  -f  3.4.5.6  = ^-— ^. 

5 

If  w  =:  9  and  5  =  3, 

1.2.3  +  2.3.4  +  3.4.5  +  4.5.6  +  5.6.7  +  6.7.8  =.  ?1^-. 
Prove  these  equations  by  computing  both  members. 
261.  Another  curious  example  is  the  following: 

Let  us  have  p  +  q  things  divided  into  two  sets,  the  one 
containing  p  and  the  other  q  things.  Then,  to  form  all  possi- 
ble combinations  of  s  things  out  of  the  whole  p  -{-  q,  we  may 
take  : 

Any  s  things  in  set  p  ; 

Or  any  combination  of  5  —  1  things  in  set^  with  any  one 
thing  of  set  ^ ; 

Or  any  combination  of  s  —  2  things  in  set  p  with  any  com- 
bination of  2  things  in  q  ; 

Or  any  combination  of  5  —  3  things  in  p  with  any  3  out 
of  g,  etc. 

We  shall  at  length  come  to  the  combinations  of  all  5  things 
out  of  q  alone.  Adding  up  these  separate  classes,  we  shall 
have : 

c?  +  cu  ci+cua +  ....  +  C1  CU  +  Gl 

This  sum  makes  up  all  combinations  of  s  things  in  the 
whole  p-\-q,  and  is  therefore  equal  to  6'^^^.  Putting  the 
numerical  expressions  for  the  combinations,  we  have  the 
theorem : 


292  COMBINATIONS. 

p-^')=©-(A)(f)-(A)©-- 

If,  as  an  example,  we  put  5  =  3,  ^  =  4,  g  —  5,  tliis  theo 
rem  will  give 

9 ^7  _  4.3»2       4^5       4  5 -^       5-4.3 
1-2.3  ~  1.2.3  "^  1-2*1  "^  1*1.2  "^  1-2^^ 

the  correctness  of  which  is  easily  proved  by  computation. 

EXERCISES. 

1.  Write  all  the  combinations  of  three  letters  out  of  the 
five,  a,  Z>,  c,  d,  e,  and  show  that  C^  of  them  begin  with  a,  CI 
with  h,  and  0%  with  c,  according  to  the  reasoning  of  §  260. 

2.  Prove  that  C|  =  6^t  +  C^, 

Cl  =  Cl  +  CI, 

and  in  general,  C^     =  C^  -{-  C^s-i- 

In  the  following  two  ways : 

(1.)  Let  all  combinations  of  s  letters  in  the  n  letters 

a,  b,  c,  ,  .  .  .  n 

be  formed,  their  number  being  (7^.  Then  suppose  one  letter 
added,  making  the  number  n  -\-  1.  The  combinations  of  s 
letters  out  of  these  n  -\-  1  will  include  the  Cg  formed  from 
the  ^  letters,  plus  each  combination  of  the  additional  (w  +  1)*^ 
letter  with  the  combinations  of  ^  —  1  out  of  the  first  n  letters. 

(2.)  Prove  the  same  general  result  from  the  formula, 


c^  =  (g. 


3.  If  we  form  all  combinations  of  3  things  out  of  7,  how 
many  of  these  combinations  will  contain  a  7,  and  how  many 
will  not  ? 

4.  If  we  form  all  the  combinations  of  s  letters  out  of  the  n 
letters 

a,  h,  c,  .  ,  ,  ,  n, 


COMBINATIONS.  293 

how  many  of  these  combinations  will  contain  a,  and  how  many 
will  not  ? 

5.  In  the  preceding  case,  how  many  of  the  combinations 
will  contain  the  three  letters  a,  d,  c  ? 

363.  Theoeem  I.  The  total  number  of  comhination$ 
ivhich  ean  he  formed  from  n  things,  including  1  zero 
combination,  is  2^.  . 

In  the  language  of  Algebra, 

c^  +  c^+c^ +  ....  +  Cl-t  -\-cl  =  t. 

Proof,  Let  ns  begin  with  3  things,  a,  h,  c,  and  let  us  call 
the  formal  zero  combination,  1  =  C^,     Then  we  have 

d,  blank,  Number  =  1 

(7i,  a,d,c.  "        =3 

C\,  ah,  ac,  be,  "        =3 

Cl,  abc,  "        =J_ 

Sum  m  8  =  23. 

Now  introduce  a  fourth  letter  d.  The  combinations  out  of 
the  four  things,  a,  h,  c,  d,  will  consist  of  the  above  8,  plus  the 
8  additional  ones  formed  by  writing  d  after  each  of  the  above 
eight.     Their  number  will  therefore  be  16. 

In  the  same  way,  it  may  be  shown  that  we  double  the  pos- 
sible number  of  combinations  for  every  thing  we  add  to  the 
set  from  which  they  are  taken.     We  have  found,  for 

n  =  3,  Sum  of  combinations  =      8  =  2^; 

71  =z  4.,  "  "  =2-8  =  24; 

n  —  ^,  "  "  =  2-24=  25; 

etc.  etc. 

which  shows  the  theorem  to  be  general. 

Theorem  II.  If  the  signs  of  the  alternate  combina- 
tions of  n  things  be  changed,  the  algebraic  sum  will  be 
zero. 

In  algebraic  language, 

Cl  _  Cl  -^Ct-Cli-  etc.  ±  Cl  ^  0.  {a) 


294  COMBINATIONS. 

Proof.    If  in  the  formula  of  §  261,  Ex.  2,  Dumely, 

^  S  ^  S    -T    ^  8-ly 

we  put  n  —  1  for  n,  it  becomes 

pn  pn~\    ,     /^^~} 

L/  s    —    ^  S         "T"    ^  s—1* 

Putting  s  successiyely  equal  to  0,  1,  2,  ...  .  n,  we  have 

ri^  p^''   1   . 

0  0   —    f   7i   —    i  ? 

Cl—    Co        +Ci        =1+Gl       , 

C  3=02+03; 

O  w-1  —    O  7i-2  +   O/^-i   —    O  7i_3  +  1. 

Substituting  these  values  in  the  expression  (a),  it  becomes 

1  _  (1  +  or')  +  {cr'  +  en  -  (er'  +  or')  + . . . . 
^  1  _  1  _  cr'  +  cr'  +  c^r'  -  or'  -  cr'  +  etc. 

How  far  soever  we  carry  this  process,  all  the  terms  cancel 
each  other  except  the  last.  Therefore,  if  we  con  tin  ae  the  addi- 
tions and  subtractions  until  we  come  to  Cn-i  ?  the  sum  will  be 

Cl  -C'ii-Cl-  etc ±  CU  =  ±  OVi  =±1. 

The  last  term  will  be  +  C'JI;  ===  =F  1?  and  will  therefore 
just  cancel  the  sum  of  the  preceding  terms. 

Note.  Theorem  I  may  be  demonstrated  by  these  same  formulae, 
since  the  sum  of  all  the  terms  taken  positively  will  be  duplicated  every 
time  we  increase  ti  by  1. 

363.  Independent  Comlinations.  There  is  a  system  of 
combinations  formed  in  the  following  Avay  : 

It  is  required  to  form  a  comhinatioTi  of  s  things,  by 
taking  one  out  of  each  of  s  different  collections.  How 
many  combinations  can  be  formed'  ? 

Let  the  1st  collection  contain  a  things, 

u       2d  ''  "       b      " 

"       3d  "  ''       c'    " 

etc.  etc. 


C0MBmATI0N8.  295 

Then  we  may  take  any  one  of  a  things  from  the  first  col- 
lection. 

With  each  of  these  we  may  combine  any  one  of  the  h  things 
in  the  second  collection. 

With  each  of  these  we  may  combine  any  one  of  the  c  things 
of  the  third  collection. 

Continuing  the  reasoning,  we  see  that  the  total  number  of 
combinations  is  the  continued  product 

ahc  ....  to  cS  factors. 

If  the  number  in  each  collection  is  equal,  and  we  call  it  a, 
the  number  of  combinations  will  be  a^. 

This  form  of  combinations  is  that  which  corresponds  most 
nearly  to  the  events  of  life,  and  is  applicable  to  many  questions 
concerning  probabilities.  For  example,  if  any  one  of  live  dif- 
ferent events  might  occur  to  a  person  every  day,  the  number 
of  diiferent  ways  in  which  his  history  during  a  year  might  turn 
out  is  52^,  a  number  so  enormous  that  255  digits  would  be  re- 
quired to  express  it. 

EXERCISES. 

1.  A  man  driving  a  span  of  horses  can  choose  one  from  a 
stud  of  10  horses,  and  the  other  from  a  stud  of  12.  How 
many  different  span  can  he  form  ? 

2.  It  is  said  that  in  a  general  examination  of  the  public 
schools  of  a  county,  the  pupils  spelt  the  word  scliolar  in  230 
different  ways.     If  in  spelling  they  might  replace 

ch  hj  c  ov  Tc\ 

0  by  au,  mu,  or  00  \ 

1  by  II', 

a  by  e,  0,  u,  or  ou; 
r  hj  re; 
in  how  many  different  ways  might  the  word  be  spelt  ? 

3.  If  a  coin  is  thrown  n  times  in  succession,  in  how  many 
different  ways  may  the  throws  turn  out  ? 

4.  If  there  are  three  routes  between  each  successive  two  of 
the  five  cities,  Boston,  New  York,  Philadelphia,  Baltimore, 
Washington,  by  how  many  routes  could  we  travel  from.  Boston 
to  Washington  ? 


296  COMBmATIONS. 

The  Binomial  Theorem  when  the  Power  is  a 
Whole  Number. 

364.  The  binomial  theorem  (§  172),  when  the  power  is  a 
positive  integer,  can  be  demonstrated  by  the  doctrine  of  com- 
binations, as  follows : 

Let  it  first  he  required  to  form  the  product  of  the  n 
hinomial  factors, 

To  understand  the  form  of  the  product,  let  us  first  study  the  special 
case  when  /?.  =  3.  Performing  the  multiplication  of  the  first  three  fac- 
tors, the  product  will  consist  of  eight  terms : 


'     ^2'^'i^3    ~T   (t^X^X-2    -f-   X^X^X^ 

This  product  is  the  expression  {a)  developed  when  7i  =  3. 


IS''?* 


We  conclude,  by  induction,  that  the  entire  product  (a) 
when  developed  in  this  same  way,  will  be  composed  of  a  sum 
of  terms,  each  term  being  a  product  of  several  literal  factors. 

When  (a)  is  thus  multiplied  out,  we  shall  call  the  result 
the  developed  expression. 

The  developed  expression  has  the  following  properties  : 

I.  Each  term  contains  n  literal  factors,  a's  and  x's, 
and  no  more. 

For,  suppose  x^r=i  a^,  x^  ^  a^,  to  Xn  —  cfn-  Then  the 
expression  (a)  will  reduce  to 

%^a^a^a^  ,  ,  ,  ,  an,  {V) 

and  the  developed  expression  must  assume  the  same  value ; 
that  is,  it  must  consist  of  terms  each  of  which  reduces  to  the 
expression 

a^a^a^  ....  an,  {c) 

when  we  change  x  into  a.  Now  if  it  contained  any  term  with 
either  more  or  less  than  n  factors,  it  could  not  assume  this 
form. 


THE  BINOMIAL    THEOREM.  297 

II.  The  factors  of  each  term  have  all  the  n  indices 

1,  2,  3, n. 

For,  the  index  figure  of  no  term  is  altered  by  changing  x 
into  a^  as  in  I.  Hence,  if  in  any  term  any  index  figure  were 
missing  or  repeated,  that  term  would  not  reduce  to  the  form 
(c),  whence  there  can  be  neither  omission  nor  repetition  of 
any  index. 

III.  Because  each  term  has  n  factors^  it  Tivust  either 
have 

n  factors  a; 

n  —  1  factors  a  and  one  factor  x; 
n  —  2  factors  a  and  two  factors  x ; 
In  general,  a  term  may  ha^ve  the  factor  a  repeated 
n  —  i  times,  and  x  repeated  i  times. 

IV.  In  a  term  which  contains  i  factors  x,  these  /  factors 
must  be  affected  with  some  combination  of  i  indices  out  of  the 
whole  number  1,  2,  3,  ....  ^ ;  and  the  n  —  i  «'s  must  be 
affected  by  the  complementary  combination  of  n  —  i  indices. 
We  next  inquire  whether  there  is  a  term  corresponding  to 
every  such  combination.     Let 

1,  3,  4,  7, ... . 
be  any  combination  of  i  indices,  and 

2,  5,  6,  8, ...  , 

the  complementary  combination  of  ^  —  /  indices. 

Since  the  developed  expression  must  be  true  for  all  values 
of  a  and  x,  let  us  put  in  (^), 

a^  ■-—  0,  ccg  =  0 ; 

«3  =  %         ^'6  =  ^; 

«4  =  0,  x^  —^\  {d) 

a^  =0,  x^  z=^  Q'y 

etc.  etc. 

The  product  {a)  will  then  reduce  to  the  single  term, 

x^a^x^x^a^a^x^a^ {e) 

By  the  same  change  the  developed  expression  must  reduce 
to  this  same  value,  and  it  cannot  do  this  unless  the  expression 
(c)  is  one  of  its  terms. 


298  GOMBmATIONS. 

Hence  the  developed  expression  must  contain  a  term 
corresponding  to  every  combination. 

V.  Since  every  combination  of  i  figures  out  of  1,  2,  3, ....  :^ 
will,  in  this  way,  give  rise  to  a  term  like  (e),  containing  the 
symbol  a  i  times,  and  the  symbol  x  n  —  i  times,  there  will  be 
C'i  such  terms. 

Now  suppose      a^  =1  a^  :=  a^  =1  ,  ,  ,  ,  cin  =^  ci' 

/y         —    /y         /y         —  /v»         /y 

*t/ -|     -^—    «^2     "    '^  ^     ~~~~     •    •    •    •    *^n     «^» 

The  expression  (a)  will  then  reduce  to  {a  -f  x)^. 

In  the  developed  expression,  all  the  Cf  terms  containing  x 

i  times  and  a  n  —  i  times  will  now  be  equal  and  their  sum 

will  reduce  to  0^  a'~^  x. 

Hence,  putting  in  succession  i  =  0,  ^  =  1,  etc.,  to  i  =  n, 

we  shall  have 

{a-\-xY  =  a^+  Cia'^-^x+C^a'^'^x^  + +Cl  tax^-^-^:^. 

Substituting  for  Ct  its  value,  we  shall  have 

{a  +  xY  =  a«  +  naP'-'^x  +  (^L^-2^2  + . . . .  +  ( — ^  Xix'^-'^  +  l-jx'^, 

which  is  the  Binomial  Theorem,  enunciated,  but  not  demon- 
strated, in  Book  V,  Chapter  I. 

Note.  If  the  studeAt  has  any  difficulty  in  understanding  the  steps 
of  the  preceding  demonstration,  he  should  suppose  71  =  3,  and  refer  the 
demonstration  to  the  developed  expression  (aO. 


PM0BABILITIE8,  -   299 

CHAPTER    III. 

THEORY     OF     PROBABILITIES. 

365.  Bef,  The  Theory  of  Probabilities  treats  of 
the  chances  of  the  occurrence  of  events  which  cannot  be 
foreseen  with  certainty. 

Notation,     Let  a  bag  contahi  4  balls,  of  which  1  is  white- 
and  3  black.    If  a  ball  be  drawn  at  random  from  the  bag,  we 
should,  in  ordinary  language,  say  that  the  chances  were  1  to  3 
in  favor  of  the  ball  being  white,  or  3  to  1  in  favor  of  its  being 
black. 

In  the  language  of  probabilities  we  say  that  the  probability 

1  3 

of  a  white  ball  is  -r.  and  that  of  a  black  one  -• 
4^  4 

In  general,  if  there  are  m  chances  in  favor  of  an  event,  and 
n  chances  against  it,  its  probability  is Hence, 

Bef,  The  Probability  of  an  event  is  the  ratio  of 
the  chances  which  favor  it  to  the  whole  number  of 
chances  for  and  against  it. 

Illustrations,     If  an  event  is  certain,  its  probability  is  1. 
If  the  chances  for  and  against  an  event  are  even,  its  prob- 
ability is  --• 

If  an  event  is  impossible,  its  probability  is  0. 

Cor,  1.  If  the  probability  that  an  event  will  occur 
is^,  the  probability  that  it  will  fail  is  1—p, 

Cor,  2.  A  probability  is  always  a  positive  fraction, 
greater  than  0  and  less  than  1. 

266.  Method  of  ProbaMUties,  To  find  the  probability  of 
an  event,  we  must  be  able  to  do  two  things : 


300  PnOBABILlTIES. 

1.  Enui7ierate  all  possible  ways  in  which  the  event 
may  occur  or  fail,  it  being  supposed  that  these  zuays 

are  all  equally  probable. 

2,  Determine  how  many  of  these  zuays  will  lead  to 
the  event. 

If  n  be  the  total  number  of  ways,  and  m  the  number  which 
lead  to  the  event,  the  probability  required  is  —  • 

EXERCISES. 

1.  A  die  has  2  white  and  4  black  sides.  What  is  the  prob- 
ability of  throwing  a  whifce  side  ? 

2.  A  bag  contains  n  balls  numbered  from  1  to  n,  the  even 
numbers  being  white  and  the  odd  ones  black.  What  is  the 
probability  of  drawing  a  black  ball  when  n  is  an  odd  number? 
What,  when  n  is  an  even  number  ? 

3.  A  bag  contains  3w-f-2  balls,  of  which  numbers  1,  4,  7, 
etc.,  are  white;  2,  5,  8,  etc.,  are  red;  3,  6,  9,  etc.,  are  black. 
What  are  the  respective  probabilities  of  drawing  a  white,  red, 
and  black  ball  ?  ,  . 

Rem.    In  tlie  last  example  tlie  probabilities  are  all  less  than  ^  ;  tli^re- 

fore,  should  one  attempt  to  guess  the  color  of  the  ball  to  be  drawn,  he 
would  be  more  likely  to  be  wrong  than  right,  no  matter  what  color  he 
guessed.  This  exemplifies  a  lesson  in  practical  judgment  to  be  drawn 
from  the  theory  of  probabilities.  If  there  are  three  or  more  possible  re- 
sults of  any  cause,  it  may  happen  that  the  best  judgment  would  be  more 
likely  to  be  wrong  than  right  in  attempting  to  predict  the  result.  Thus, 
if  there  are  three  presidential  candidates  with  nearly  equal  chances,  the 
chances  would  be  against  the  election  of  any  one  that  might  be  named. 

Gamblers  of  the  turf  are  nearly  always  found  betting  odds  against 
every  horse  that  may  be  entered  for  a  race,  though  it  is  certain  that  one 
of  them  will  win. 

Hence,  if  a  naturaj.  event  may  arise  from  a  number  of  causes  with 
nearly  equal  facility,  it  is  unphilosophical  to  have  any  theory  whatever 
of  the  cause,  because  the  chances  may  be  against  the  most  probable 
cause  being  the  true  one. 

Probabilities  depending:  upoi?!  Combinations. 
267.     Problem  i.    Two  coins  are  thrown.    What  are  the 
respective  probabilities  that  the  result  will  be  :    Both  heads  ? 
head  and  tail  ?  both  tails  ? 


PnOBABlLITIES,  301 

At  first  sight  it  might  appear  that  the  chances  in  favor  of 
these  three  results  were  equal,  and  that  therefore  the  probabil- 
ity of  each  was  ^-    But  this  would  be  a  mistake.     To  find  the 

probabilities,  we  must  combine  the  possible  throws  of  the  first 

coin  (which  call  A)  with  the  possible  throws  of  the  second 
(which  call  B),  thus  : 

A,  head  ;  B,  head. 

.   A;  head ;  B,  tail. 

A,  tail ;  B,  head. 

A,  tail ;  B,  tail. 

These  combinations  are  all  equally  probable,  and  while 

there  are  only  one  each  for  both  heads  and  both  tailsVthere  are 

1^1     1 
two  for  head  and  tail.     Hence  the  probabilities  are  7,  -,  ^^ 

The  sum  of  these  three  probabilities  is  1,  as  it  ought  always 
to  be  when  all  possible  results  are  considered. 

Proh.  2.  Five  coins  are  thrown.  What  are  the  respective 
probabilities:         q  heads,  5  tails? 

1  head,  4  tails?  .^ m-  ^,^^ 

2  heads,  3  tails? 
etc.  etc. 

Let  the  several  coins  be  marked  a,  1),  c,  d,  e.  Coin  a  may 
be  either  head  or  tail,  making  two  cases.  Each  of  these  two 
cases  of  coin  a  may  be  combined  with  either  case  of  b  (as  in  the 
last  example),  making  4  cases. 

Each  of  these  4  cases  may  be  combined  with  either  case  of 
coin  c,  making  8  cases. 

Continuing  the  process,  the  total  number  of  cases  for  five 
coins  is  5^  ±=  32. 

Of  the^e  32  cases,  only  one  gives  no  head  and  5  tails. 

There  fare  5  cases  of  1  head,  namely:  a  alone  head,  I?  alone 
head,  etc.,'  to  e. 

2  heads  may  be  th  ^wn  by  coins  r/,  b;  a,  c,  etc. ;  b,  c  ;  b,  d, 
etc. ;  c,  d,  etc. ;  that  is,  by  any  combination  of  two  letters  out 
of  the  five,  a,  b,  c,  d,  e.     Hence  the  number  of  cases  is 

Ol  =  10. 


302  PBOBABILiriES. 

In  the  same  way  the  number  of  cases  corresponding  to  3, 
4,  and  5  heads  are,  respectively, 

Gl  =  10,         Cl  =  6,         Cl  =  1. 
Dividing  by  the  whole  number  of  cases,  we  find  the  respec- 
tive probabilities  to  be 

32'         32'     32'         32'     32'     32* 

The  following  general  proposition  is  now  to  be  proved  by 
the  student : 

Theorem.  If  there  are  n  coins,  the  prohahility  of 
throwing  s  heads  and  n  —  s  tails  is 

From  this  result  we  may  prove  the  theorem  in  combina- 
tions of  §  262.  If  we  suppose,  in  succession,  s  =  0,  8  =  1, 
s  =:  2,  etc.,  to  s  =:  H,  the  respective  probabilities  of  0  iiead, 
1  head,  2  heads,  etc.,  will  be 

'     Cl      G^      Cl  '     -     Cl 

2^'     2^'     2^*'  2^' 

Because  the  sum  of  all  these  probabilities  must  be  unity, 
we  find 

C^  +   ^5l   +   C'l  +  ....    +    a;:  =::   2^ 

Prol,  3.  Two  dice  are  thrown  at  backgammon.  What  are 
the  respective  probabilities  of  throwing  5  and  6  and  two  6's  ? 

If  we  call  the  dice  a  and  h,  any  number  from  1  to  6  on  « 
may  be  combined  with  any  number  from  1  to  6  on  Z>.  There- 
fore, there  are  in  all  36  possible  combinations. 

In  order  to  throw  two  6's,  a  must  come  6  and  h  also. 
Therefore  there  is  only  one  case  for  this  result,  so   that  its 

probability  is  --• 

To  bring  5  and  6,  a  may  be  5  and  Z>  6,  or  ^  5  and  a  6.  So 
there  are  two  cases  leading  to  this  result,  and  its  probability  is 

A  _  _L 
•'  36  ""^  18' 


PROBABILITIES.  303 

Note.  That  5  and  6  are  twice  as  probable  as  a  double  6  may  be 
clearly  seen  by  supposing  that  the  two  dice  are  thrown  in  succession.  If 
the  first  throw  is. either  5  or  6,  there  is  a  chance  for  the  combination  5,  6, 
but  there  is  no  chance  for  a  double  6  unless  the  first  throw  is  6. 

Proh  4.    If  three  dice  are  thrown,  what  are  the  respective 

probabihties  that  the  numbers  will  be : 

1,  1,  1?  1,  1,  2?  1,  2,  3? 

The  solution  of  this  case  is  left  as  an  exercise  for  the 
student. 

Proh.  5.  From  a  bag  containing  3  white  and  2  black  balls, 
2  balls  are  drawn.     What  are  the  respective  probabilities  of 

Both  balls  white? 
1  white  and  1  black  ? 
Both  black? 

Since  any  2  balls  out  of  5  may  be  drawn,  the  total  number 
of  cases  is  Cg. 

Only  one  of  these  combinations  consists  of  two  white  balls. 

C\  of  the  cases  bring  both  balls  black. 

A  w^hite  and  black  are  formed  by  combining  any  one  of  the 
three  white  with  any  one  of  the  two  black. 

The  respective  probabilities  can  now  be  deduced  by  the 
student. 

EXERCISES. 

1.  It  takes  two  keys  to  unlock  a  safe.  They  are  on  a 
bunch  with  two  others.  The  clerk  takes  three  keys  at  random 
from  the  bunch.  What  is  the  probability  that  he  has  both  the 
safe  keys? 

2.  A  party  of  three  persons,  of  whom  two  are  brothers,  seat 
themselves  at  random  on  a  bench.  What  are  the  probabilities 
{a)  that  the  brothers  will  sit  together,  {h)  tliat  they  will  have 
the  third  man  between  them  ? 

3.  If  two  dice  are  thrown  at  backgammon,  what  are  the 

probabilities 

{a)  Of  two  aces  ? 

{h)  Of  one  ace  and  no  more  ? 

4.  In  order  that  a  player  at  backgammon  may  strike  a  cer« 


304  PROBABILITIES. 

tain  point,  the  sum  of  the  numbers  thrown  must  be  8.     What 
are  his  cliances  of  succeeding  in  one  throw  of  his  two  dice  ? 

5.  A  purty  of  13  persons  sit  at  a  round  table.  What  is  the 
probability  that  Mr.  Taylor  and  Mr.  Williams  will  be  next  to 
each  other?     (See  §  253.) 

6.  An  illiterate  servant  puts  two  works  of  2  volumes  each 
upon  a  shelf  at  random.  What  is  the  probability  that  both 
pair  of  companion  volumes  are  together? 

7.  A  gentleman  having  three  pair  of  boots  in  a  closet^  sent 
a  blind  valet  to  bring  him  a  pair.  The  valet  took  two  boots  at 
random.  What  are  the  chances  that  one  was  right  and  tlie 
other  left  ?    What  is  the  probability  that  they  were  one  pair  ? 

8.  If  the  volumes  of  a  3-  volume  book  are  placed  at  random 
on  a  shelf,  what  is  the  probability  that  they  will  be  in  regular 
order  in  either  direction  ? 

9.  A  man  wants  a  particular  span  of  horses  from  a  stud 
of  8.  His  groom  brings  him  5  horses  taken  at  random.  What 
is  the  probability  that  both  horses  of  the  span  are  amongst 
them  ? 

10.  From  a  box  containing  5  tickets,  numbered  1  to  5, 
3  are  drawn  at  random.  What  is  the  probability  that  numbers 
2  and  5  are  both  amongst  them  ? 

1 1.  The  same  thing  being  supposed,  what  is  the  probability 
that  the  sum  of  the  two  numbers  remaining  in  the  box  is  6  ? 

12.  Of  two  purses,  one  contains  5  eagles  and  another  10 
dollar-pieces.  If  one  of  the  purses  is  selected  at  random,  and 
a  coin  taken  from  it,  what  is  the  probability  that  it  is  an 
eagle  ? 

13.  From  a  bag  containing  3  white  and  4  black  balls 
2  balls  are  drawn.  What  is  tlie  probability  that  they  are  of 
the  same  color  ? 

14.  The  better  of  two  chess  players  is  twice  as  likely  to  win 
as  to  be  beaten  in  any  one  game.  What  chance  has  his  weaker 
opponent  of  winning  2  games  in  a  mateli  of  3  ? 

15.  From  a  bag  containing  m  white  and  n  black  balls,  two 
balls  are  drawn  at  random.  What  is  the  probability  that  oii'.' 
is  white  and  the  other  black  ? 


PROBABILITIES.  •  305 

1 6.  From  a  bag  containing  1  white,  2  red,  and  3  black 
balls,  3  balls  are  drawn.  What  is  the  probability  that  they  are 
all  of  different  colors  ? 

17.  If  ^  coins  are  thrown,  what  is  the  chance  that  there 
will  be  one  head  and  no  more  ? 

18.  From  a  Congressional  committee  of  6  Republicans  and 
5  Democrats,  a  sub-committee  of  3  is  chosen  by  lot.  What  is 
the  probability  that  it  will  be  composed  of  two  Eepublicans 
and  one  Democrat  ? 

Compound  Events. 

368.  Theorem  I.  The  probability  that  tiuo  independ- 
ent events  will  both  happen  is  equal  to  the  product  of 
their  separate  probabilities. 

Proof,  For  the  first  event  let  there  be  m  cases,  of  which 
p  are  favorable;  and  for  the  second  ?2  cases,  of  which  ^  are 
favorable.     Then,  by  definition,  the  respective   probabilities 

will  be  —  and  -  • 
m  n 

When  both  events  are  tried,  any  one  of  the  in  cases  may  be 
combined  with  any  one  of  the  n  cases,  making  in  all  in  x  n 
combinations  of  equal  probability. 

The  combinations  favorable  to  both  events  will  be  those 
only  in  which  one  of  the  p  cases  favorable  to  the  first  is  com- 
bined with  one  of  the  q  cases  favorable  to  the  second.  The 
number  of  these  combinations  is  p  x  q. 

Therefore  the  probability  that  both  events  will  happen  is 

P  ><  q  ^  Z  X  2 
m  X  n       m      n^ 

which  is  the  product  of  the  individual  probabilities. 

If  there  are  three  events  of  which  the  probabilities  are  ^,  q^ 
and  r,  and  we  wish  to  find  the  probability  that  all  three  will 
happen,  we  may  by  what  precedes  regard  the  concumng  of  the 
first  two  events  as  a  single  event,  of  which  the  probitbility  is 
pq.  Then  the  probability  that  the  third  event  will  also  con- 
cur is  the  product  of  this  probability  into  r,  or 

pqr. 
20 


306  PROBABILITIES. 

Proceeding  in  the  same  way  with  4,  5,  6,  ...  .  events,  we 
reach  the  general 

Theorem  II.  The  -probability  that  any  number  of  in^ 
dependent  events  will  all  occur  is  equal  to  the  continued 
product  of  their  individual  probabilities. 

Eem.  This  theorem  is  of  great  practical  use  as  a  guide  tcj 
our  expectations.  It  teaches  that  if  success  in  an  enterpr 
requires  the  concurrence  of  a  great  number  of  favorable  cil 
cumstances,  the  chances  may  be  greatly  against  it,  although 
each  circumstance  is  more  likely  than  not  to  occur. 

This  is  illustrated  by  the  following 

Example  i.  A  traveller  on  a  journey  by  rail  has  8  connec- 
tions to  make,  in  order  that  he  may  go  through  on  time. 
There  are  two  chances  to  one  in  favor  of  each  connection. 
What  is  the  probability  of  his  keeping  on  time  ? 

The  probability  of  each  connection  being  - ,  the  probabil- 

o 

ity  of  successfully  making  the  first  two  connections  will,  by  the 
preceding  theorems,  be  ( - 1 ,  the  first  three  I  - 1 ,  and  all  eight 


/2\8  _  28 
\3/    ~"  38 


28        256  1 

3-8==  6561  ^2:6'^^"^^^- 


Therefore  there  are  25  chances  to  1  against  his  going- 
through  on  time. 

On  the  other  hand,  if,  instead  of  any  one  accident  being 
fatal  to  success,  success  can  be  prevented  only  by  the  concur- 
rence of  a  series  of  accidents,  the  probability  of  failure  may 
become  very  small. 

Ex.  2.     A  ship  starts  on  a  voyage.    It  is  an  even  chance 

that  she  will  encounter  a  heavy  gale.    The  probability  that 

9 
she  will  not  spring  a  leak  in  the  gale  is  ^n*     ^^  ^  ^^^^  occurs, 

9 
there  is  a  probability  of  —  that  the  engine  will  be  able  to 

3 
pump  her  out.     If  they  fail,  the  probability  is  j  that  the  com- 


PROBABILITIES,  307 

partments  will  keep  the  ship  afloat.  If  she  sinks,  it  is  an  even 
chance  that  any  one  passenger  will  be  saved  by  the  boats 
What  is  the  probability  that  any  individual  passenger  will  be 
lost  at  sea  ? 

The  probability  that 

the  ship  will  meet  a  heavy  gale  is ^ 

the  ship  will  spring  a  leak  in  the  gale  is To 

the  engines  cannot  pump  her  out  is    .    . — 

the  compartments  cannot  keep  her  afloat  is - 

the  boats  cannot  save  the  passenger  is     .    : - 

The  continued  product  of  these  probabilities  is  ^Tuio^ 
which  is  the  probability  that  the  passenger  will  be  lost. 

369.  The  preceding  theorem  as  enunciated  supposes  that 
the  several  events  are  independent,  that  is,  that  the  probability 
of  the  occurrence  of  any  one  is  not  affected  by  the  occurrence 
or  non-occurrence  of  the  others.  To  investigate  what  modifi- 
cation is  required  when  the  occurrence  of  one  of  the  events 
alters  the  probabihty  of  another  of  the  events,  let  us  distinguish 
the  two  events  as  i\\Q  first  and  second.     We  then  reason  thus : 

Let  the  total  number  of  equally  possible  cases  be  m,  and  let 
p  of  these  cases  favor  the  first  event.     Its  probability  will 

then  be  — • 
ra 

It  is  certain  that  the  events  cannot  both  happen  unless  the 
first  one  happens.  Hence  the  cases  which  favor  both  events 
can  be  found  only  among  the  p  cases  which  favor  the  first. 
Let  q  of  these  p  cases  favor  the  second  event.  Then  the  prob- 
ability of  both  events  will  be  —  • 

In  case  the  first  event  happens,  one  of  the  p  cases  which 


308  PROBABILITIES. 

favor  it  must  occur,  aud  the  probability  of  the  second  event 

will  then  be  ^.    Then 
P 

Probability  of  both  events  =  —  =  —  x  -•    Hence, 
•^  m       7)1      p 

Theokem.  The  probability  that  two  events  will  both 
oceur  is  equal  to  the  probability  of  the  first  event  mitlti- 
pliecl  by  the  probability  of  the  seeond,  in  case  the  first 
occurs. 

By  continuing  the  reasoning  to  more  events,  we  reach  the 
general 

Theokem.  The  probability  that  a  nuT)%ber  of  events 
will  all  occur  is  equal  to  the  product 

{X  Prob.  of  second  in  case  first  occurs. 
X  Prob.  of  third  in-  case  first  two  occur. 
X  Prob.  of  fourth  in  case  first  three  occur, 
etc.  etc.  etc. 

Example.  From  a  bag  containing  2  white  and  3  black 
balls,  2  balls  are  drawn.  What  are  the  probabilities  (1)  that 
both  balls  are  white,  (2)  that  both  are  black  ? 

This  problem  has  already  been  solved,  but  we  are  now  to 
see  how  the  answers  may  be  reached  by  the  last  theorem.  It 
is  evident  that  we  may  suppose  the  two  balls  drawn  out  one 
after  the  other,  and  the  probabilities  of  their  being  white  or 
black  will  be  the  same  as  if.  both  were  drawn  together. 

I.  Both  balls  white.     The  probability  that  the  first  ball 

2 
drawn  is  white  is  -•    If  it  really  proves  to  be  white,  there  will 
o 

be  left  1  white  and  3  black  balls.     In  this  event,  the  probabiliti^ 

that  the  second  also  will  be  white  is  - 

4 

Hence  the  probability  that  both  are  white  is 

2       1  _  2. 

5  ^  4  ""  10* 


PROBABILITIES,  309 

II.  Both  halls  hlach.    Applying  the  same  reasoning,  we 
find  for  the  probability  of  this  case, 
3       1  _  ^ 
5  '"^  2  ~"  10* 

EXERCISES. 

I.  Two  men  embark  in  separate  commercial  enterprises. 
The  odds  in  favor  of  one  are  3  to  2 ;  in  favor  of  the  other,  2 
to  1.  What  are  the  probabilities  (1)  that  both  will  succeed  ? 
(2)  that  both  will  fail? 

-2.  The  probability  that  a  man  will  die  within  ten  years  is 

-,  and  that  his  wife  will  die  is  —•     What  are  the  respective 

probabilities  that  at  the  end  of  ten  years, 

(a)  Both  are  living  ? 

(/3)  Both  are  dead  ? 

(y)  Husband  living,  but  wife  dead? 

{6)  Husband  dead,  but  wife  living  ? 

2 

3.  The  probability  that  a  certain  door  is  locked  is  -  •     The 

o 

key  is  on  a  bunch  of  4.  A  man  takes  2  of  the  four  keys,  and 
goes  to  the  door.  What  are  the  chances  that  he  will  be  able  or 
unable  to  go  through  it  ? 

4.  Two  bags  contain  each  4  black  and  3  white  balls.  A 
jDerson  draws  a  ball  at  random  from  the  first  bag,  and  if  it  be 
white  he  puts  it  into  the  second  bag,  mixes  the  balls,  and  then 
draws  a  ball  at  random.  What  is  the  probability  of  drawing 
a  white  ball  from  each  of  the  bags  ? 

5.  If  a  Senate  consists  of  m  Democrats  and  71  Eepublicans, 
what  is  the  probability  that  a  committee  of  three  will  include 
2  Democrats  and  1  Eepublican? 

6.  A  bag  contains  2  white  balls  and  5  black  ones.  Six 
people,  A,  ,B,  0,  D,  E,  F,  are  allowed  to  go  to  the  bag  in  alpha- 
betical order  and  each  take  one  ball  out  and  keep  it.  The 
first  one  who  draws  a  white  ball  is  to  receive  a  prize.  What 
are  their  respective  chances  of  winning? 

Note.  A's  chance  is  easily  calculated,  because  lie  has  the  draw  from 
all  7  balls. 


310  PROBABILITIES 

In  order  that  B  may  win,  A  must  first  fail.  Therefore,  to  find  B's 
probability  we  find  (1)  the  probability  that  A  fails,  (2)  the  probability  that 
if  A  fails  then  B  will  win.  We  then  take  the  product  of  these  probabili- 
ties. 

In  order  that  C  may  gain  the  prize,  (1)  A  must  fail,  (2)  B  must  fail, 
(3)  C  himself  must  gain.  So  we  find  the  successive  probabilities  of  these 
occurrences. 

Continuing  to  F,  we  find  that  he  cannot  win  unless  the  5  men  before 
him  all  miss.  He  is  then  certain  to  gain,  because  only  the  two  white 
balls  would  be  left. 

7.  Two  men  have  one  throw  each  of  a  coin.  X  offers  a 
prize  if  A  throws  head,  and  if  he  fails,  but  not  otherwise,  B 
may  try  for  the  prize.  If  both  fail,  X  keeps  the  prize  himself. 
What  are  the  respective  chances  of  the  three  men  having  the 
prize  ? 

8.  A  and  B  are  alternately  to  throw  a  coin  until  one  of 
them  throws  a  head  and  becomes  the  winner.  If  A  has  the 
first  throw,  what  are  their  respective  chances  of  winning  ? 

9.  A  crowd  of  71  men  are  allowed  to  throw  in  the  same  way 
for  a  prize,  in  alphabetical  order,  the  game  ceasing  as  soon  as  a 
head  is  thrown.  What  are  the  respective  chances  of  the  con- 
testants? 

10.  Three  men  take  turns  in  throwing  a  die,  and  he  who 
first  throws  a  6  wins.     What  are  their  respective  chances  ? 

11.  If  4  cards  are  drawn  from  a  pack  of  52,  show  that  the 
probability  that  there  will  be  one  of  each  of  the  four  suits  is 

39  26  13 
51*50*49' 

12.  One  purse  contains  5  dimes  and  1  dollar,  and  another 
contains  6  dimes.  5  pieces  are  taken  from  the  first  purse  and 
put  into  the  second,  and  after  being  mixed  5  are  taken  from 
the  second  and  put  into  the  first.  Which  purse  is  now  most 
likely  to  contain  the  dollar  ? 

13.  Of  two  purses,  one  contains  4  eagles  and  2  dollars,  the 
other  4  eagles  and  6  dollars.  One  being  taken  at  random,  and 
a  coin  drawn  from  it,  what  are  the  respective  probabilities 
that  it  is  an  eagle  or  a  dollar  ? 


PROBABILITIES.  311 


Cases  of  Unequal  Probability. 

370.  Def.  If  two  or  more  possible  events  are  so 
related  that  only  one  of  them  can  happen,  they  are 
called  Mutually  Exclusive  Events. 

Theorem.  The  prdbahility  that  some  one  of  several 
exclusive  events,  we  care  not  which,  ivill  occur,  is  equal 
to  the  sum  of  their  separate  probahilities. 

Proof,  Let  there  be  m  possible  and  equally  probable  eases 
in  all;  let  p  of  these  cases  be  favorable  to  one  event,  q  to  the 

P       Q        T 

second,  r  to  the  third,  etc.,  so  that  — ,  — .,  — ,  are  the  re- 

,,.,.,.  m     m    m 

spective  probabilities. 

Since  only  one  of  the  events  is  possible,  the  p  cases  which 
favor  one  must  be  entirely  different  from  the  q  cases  which 
favor  the  second,  and  these  cases  p-\-q  must  be  entirely  differ- 
ent from  the  r  which  favor  the  third,  etc. 

Hence  there  will  be  j9  +  ^^  +  r  +  etc. ,  cases  which  favor  some 
one  or  another  of  the  events.  Hence  the  probability  that  some 
one  of  these  events  will  occur  is 

p  -\-  q  -^^  r  -\-  etc. 
m  ' 

which  is  equal  to  the  sum  of  the  probabilities, 

par, 

—  +  —  H h  etc. 

m       m      m 

Eem.  If  the  concurrence  of  some  two  events,  say  the  first 
and  second,  had  been  possible,  some  one  or  more  of  the  p  cases 
which  favor  the  first  would  have  been  found  among  the  q  cases 
which  favor  the  second.  Then  the  whole  number  of  cases 
which  favored  either  event  would  have  been  less  than  p-{-q, 
and  the  probability  that  one  of  the  two  events  would  happen 
less  than  the  sum  of  their  respective  probabilities. 

3*71.  General  Problem.  To  find  the  probahility  that 
an  event  of  ivhich  the  proiahility  on  any  one  trial  is  p, 
will  happen  exactly  s  tiines  in  n  trials. 


312  PROBABILITIES. 

This  problem  is  at  the  basis  of  some  of  the  widest  applica- 
tions of  the  theory  of  probability  to  practical  questions,  espe- 
cially those  associated  with  life  and  fire  insurance.  The  con- 
ditions which  it  implies  are  therefore  to  be  fully  comprehended. 

We  may  conceive  a  trial  to  mean  giving  the  event  an  opj^or- 
hmity  to  happen.  The  simplest  kind  of  trial  is  that  of  throw- 
ing a  coin  or  die.  At  each  throw,  any  side  has  an  opportunity 
to  come  up.  Then,  if  we  throw  50  pieces,  or  which  amounts 
to  the  same  thing,  throw  the  same  piece  50  times,  there  will 
be  50  trials;  and  we  may  inquire  into  the  probability  that  a 
given  side  will  be  thrown  exactly  9  times  in  these  trials. 

The  same  conception  occurs  in  another  form  if  we  have  50 
men,  each  of  whom  has  an  equal  chance  of  dying  within 
5  years.  Waiting  to  see  if  any  one  man  will  die  in  the  course 
of  the  5  years  is  a  trial,  so  that  there  are  50  trials  in  all,  and 
we  may  inquire  into  the  probability  that  9  of  the  men  will  die 
during  the  trials,  just  as  in  the  case  of  50  throws  of  a  die. 

Let  us  distinguish  the  several  trials  by  the  letters 
a,  h,  c,  d,  e,  ,  ,  ,  .  ii, 
which  must  be  ?^  in  number. 

1.  In  order  that  the  event  may  not  happen  at  all,  it  must 
fail  on  every  one  of  the  n  trials.  The  probability  of  this 
(§  268,  Th.  II)  is  (1  —pY.  This  is  therefore  the  probability 
that  it  will  not  happen  at  all. 

Because  the  probability  of  tlie  event  happening  on  any  one 
trial  is  }),  the  probability  of  its  failing  is  1  —  p.  We  now 
compare  the  possible  results. 

2.  The  event  may  happen  once  on  any  one  of  the  n  trials, 
«,  h^  c,  etc.  In  order  that  it  may  happen  only  once,  it  must 
fail  on  the  other  /z.  —  1  trials.  The  probability  that  it  will 
happen  on  any  one  trial,  say  e,  and  also  fail  on  the  remaining 
n  —  1  trials  is,  by  the  same  theorem, 

p  (1  —py-'^. 

Because  there  are  7i  trials  on  which  it  may  equally  happen, 
the  probability  that  it  will  happen  once  and  only  once  is 
np  (1  —  pY-\ 


PROBABILITIES,  313 

3.  The  event  may  happen  twice  on  any  two  trials  out  of  the 
n  trials.  In  order  that  it  may  happen  twice  only,  it  must  fail 
on  the  other  ^  —  2  trials.     Taking  any  one  combination,  say 

Happen  on     d,  d\ 

Fail  on  a,  c,  e,  ,  ,  .  ,  n, 

the  probability  is  p^  (1  —  pY~'^, 

But  it  may  happen  twice  on  any  combination  of  two  trials 
out  of  the  n  trials,  a,  h,  c,  ,  ,  ,  .  n.  Because  these  combina- 
tions are  mutually  exclusive  (§  270),  the  total  probability,  of 
happening  twice  is 

4.  In  general,  in  order  that  the  event  may  happen  just  s 
times,  it  must  happen  on  some  combination  of  s  trials,  and  fail 
on  the  complementary  combination  of  n  —  5  trials.  The 
probability  on  any  one  combination  is  p^  (1  —  pY~^  and  there 
are  (7?  such  combinations.  Hence  the  general  probability  of 
happening  s  times  is 

C^,ps{l-p)n-s,  {a) 

If  there  is  on  each  trial  an  equal  chance  for  and  against 
the  event,  then  p  =  -.  and  1  —p  =  --  The  probability  of 
the  event  happening  s  times  then  becomes 

2n' 

This  case  corresponds  to  that  already  treated  in  §  267, 
Problem  2,  and  the  result  is  the  same  there  found. 

HXERCISES. 

I.  A  die  having  two  sides  white  and  four  sides  black  is 
thrown  5  times.  What  are  the  respective  probabilities  of  a 
white  side  being  thrown  1,  2,  3,  4,  and  5  times? 

Note.    Here  p,  tlie  probability  of  a  white  side  on  one  throw,  is  ^ ,  and 

2 
1  —  p  =  -  •   The  number  n  of  trials  is  5. 
o 


314  PROBABILITIES. 

2.  Of  6  healthy  men  aged  50,  the  probability  that  any  one 
will  live  to  80  is  J-    What  is  the  probability  that  three  or 

more  of  them  will  live  to  that  age  ? 

3.  A  chess-player  whose  chances  of  winning  any  one  game 
from  his  opponent  are  as  2  to  1,  undertakes  to  win  3  games 
out  of  4.     What  is  the  probability  that  he  will  be  able  to  do  it? 

Note.  It  would  be  a  fallacy  to  suppose  that  the  probability  required 
is  that  of  winning  exactly  3  games,  because  he  will  equally  win  if  he 
wins  all  four  games. 

3*73.  Events  of  Maximum  ProhaUlity.    Keturning  to  the 
general  expression  {a),  let  us  inquire  what  number  of  times 
the  event  is  most  likely  to  occur  on  n  trials.     The  required 
'  number  is  that  value  of  s  for  Avhich  the  probability 

is  the  greatest. 

If  we  call  Ps  the  probability  that  the  event  will  happen 
exactly  s  times,  and  if  s  is  to  be  the  number  for  which  the 
probability  is  greatest,  we  must  have 

Pa  >   Ps-U 
Ps>  Ps^^l^ 

Substituting  for  these  quantities  the  corresponding  forms 
of  the  expression  (a),  which  is  equal  to  Ps,  we  have 

C'sP'ii  -pY-'  >  otip'-'ii  -pY-'^\ 
(7?iJ^(i  -pY-'  >  c^+tp'^'ii-pY-'-'^ 

The  general  formula  for  (7?  in  §  257  gives 


(b) 


Og    — ^  s-h 


cS    +     1 


w 


Hence  we  have,  by  dividing  both  terms  of  the  first  in- 
equality (Jb)  by  C^-ip'-^  (1  —pY~'^ 

1 p  >  1  —  /?. 


PROBABILITIES.  315 

Multiplying  by  s,  tiiis  becomes 

np  —  sjj  -\-  2^  >  ^  ~  "^i^- 
Interchanging  the  members  and  reducing,  we  have 

s  <  p  (n  +  1).  (d) 

Now  divide  the  second  inequality  (b)  by  C^  p^  (1  —  ^j)^~^^ 
and  reducing  by  the  second  equation  (c),  we  have 
-I  ^  n  —  s 

Multiplying  by  5  +  1  and  reducing,  we  find 

s  >  p  {n  +  1)  -  1.  {e) 

Comparing  the  inequalities  (d)  and  (e),  we  see  that  s  lies 
between  the  two  quantities  p  {n  -{-  1)  and  p  {n  -|-  1)  —  1 ; 
that  is, 

s  is  the  greatest  whole  number  in  p  (n  -\-  \), 

If  the  number  of  trials  7i  is  a  large  number,  and  ^  is  a  small 
fraction,  p{n  -\-  1)  and  pn  will  differ  only  by  the  fraction  p. 
We  shall  then  have,  very  nearly, 

6^  z=i  pn. 
That  is : 

Theorem  L  The  most  probable  number  of  times  thai 
an  event  will  happen  on  a  great  number  of  trials  is  the 
produet  of  the  number  of  tibials  by  the  probability  on 
eaeh  trial. 

Example.  If  a  life  insurance  company  has  GOOO  members, 
and  the  probability  that  each  member  will  live  one  year  is  on 

tlie  average  — ,  then  the  most  probable  number  of  deaths 

during  the  year  is  100. 

Eem.  It  must  not  be  supposed  that  in  this  case  the  num- 
ber of  deaths  is  likely  to  be  exactly  100,  but  only  that  they 
will  fall  somewhere  near  it. 

There  is  a  practical  rule  for  determining  what  deviation 
must  be  guarded  against,  the  demonstration  of  which  requires 
more  advanced  mathematical  methods  than  those  employed  in 
this  chapter.     It  is: 


316  PE0BAB1LITIE8. 

Theokem  II.  Deviations  from  the  most  probable  num- 
ber of  deaths,  equal  to  the  square  root  of  that  number, 
will  be  of  frequent  oeeurrenee. 

Deviations  mueh  greater  than  this  square  root  will 
be  of  infrequent  occurrence,  and  deviations  more  than 
twice  as  great  zvill  be  rare. 

Examples.  In  a  company  of  which  the  probable  annual 
number  of  deaths  is  10,  the  actual  number  will  commonly  fall 
between  10  —  VlO  and  10  +  V 10,  or  between  7  and  13.  It 
will  very  rarely  happen  that  the  number  of  deaths  is.  as  small 
as  4  or  as  large  as  16. 

If  the  company  is  so  large  that  the  most  probable  number 
of  deaths  is  100,  the  actual  number  will  commonly  fall  between 
100  —  VlOO  and  100  +  VlOO,  or  between  90  and  110. 

If  the  most  probable  number  of  deaths  is  1000,  the  actual 
number  will  commonly  range  between  968  and  1032. 

We  now  see  the  following  result  of  this  theorem: 

The  greater  the  number  of  deaths  to  be  expected,  the 
greater  will  be  the  probable  deviation,  but  the  less  ivill  be 
the  ratio  of  this  deviation  to  the  whole  number  of  deaths. 

Examples.  The  reductions  of  the  cases  just  cited  are 
shown  as  follows : 

Expected  number  Probable  Ratio  of  deviation 

of  deaths.  deviation.  to  expected  number. 

10  3  0.33 

100  10  0.10 

1000  32  '  0.03 

Application  to  Life  Insurance. 

373.  At  each  age  of  human  life  there  is  a  certain  proba- 
bility that  a  person  will  live  one  year.  This  probability  di- 
minishes as  the  person  advances  in  age. 

It  is  learned  from  observation,  on  the  principle  described  in 
the  preceding  section,  that  events  in  a  vast  number  of  trials 
are  likely  to  happen  a  number  of  times  equal  to  the  product  of 
their  probability  on  each  trial,  multiplied  by  the  number  of 
trials. 


PROBABILITIES.  317 

Therefore,  by  dividing  the  whole  number  of  times  the  event 
has  happened  by  the  whole  number  of  trials,  the  quotient  is 
the  most  probable  value  of  the  probability  on  one  trial. 

Example.  If  we  take  50,000  people  at  the  age  of  25,  and 
record  how  many  of  them  are  alive  at  the  end  of  one  year,  this 
is  making  50,000  trials  whether  a  person  of  that  age  will  live 
one  year. 

If  49,650  of  them  are  alive  at  the  end  of  the  year,  and  350 
are  dead,  we  would  conclude : 

ProbabiHty  of  living  one  year,     ....     0.993 
Probability  of  dying  within  the  year,  .    ,    0.007 

The  probability  for  all  ages  may  be  determined  by  taking  a 
great  number  of  infants,  say  100,000,  and  counting  how  many 
die  in  each  year  until  all  are  dead.  If  n  are  living  at  the  age 
y,   and  n'  at  the  age  ^  +  1,  then  the  probability  of  dying 

within  one  year  after  the  age  y  will  be ,  and  that  of 

livinsr  will  be  —  • 
^  n 

It  is  not,  however,  necessary  to  wait  through  a  lifetime  to 
reach  this  conclusion.  It  is  sufficient  to  find  from  observation 
what  proportion  of  the  people  of  each  age  die  during  any  one 
year.  Suppose,  for  instance,  that  the  census  of  a  city  is  taken, 
and  it  is  found  that  there  are  2500  persons  aged  30,  and  2000 
aged  50.  At  the  end  of  a  year  another  inquiry  is  made  to 
ascertain  how  many  are  dead.  It  is  found  that  20  of  the  30 
year  old  people,  and  30  of  the  50  year  old  people  have  died. 
This  would  show : 

At  age  30,  probability  of  dying  within  1  year  =  0.008. 
"      50,  "  "  "  "  —  0.015. 

This  saine  probability  being  obtained  for  every  year  of  life, 
the  probability  of  living  1  year  at  all  ages  would  be  known. 
Then  a  table  of  mortality  could  be  formed. 

A  table  of  mortality  starts  out  with  any  arbitrary  number 
of  people,  generally  100,000,  at  a  certain  age,  frequently  10 
years.  It  then  shows  how  many  of  these  people  will  be  living 
at  the  end  of  each  subsequent  year  until  all  are  dead.  The 
following  is  a  specimen  of  such  a  table. 


318 


PROBABILITIES. 


Table  of  Mortality. 


Ages. 

Living. 

Dying. 

Prob.  of 
surviving 

Prob.  of 
within 

Ages. 

Living. 

Dying. 

Prob.  of 
surviving 

Prob.  of 

dying 
within 

10 

442 

a  year. 

the  year. 

a  year. 

the  year. 

1 00000 

.99558 

.00442 

60 

58373 

1677 

.97127 

.02872 

II 

99558 

407 

.99591 

.00408 

61 

56696 

1760 

.96895 

.o3io4 

12 

99i5i 

385 

.99611 

.00388 

62 

54936 

1849 

.96634 

.03365 

i3 

98766 

376 

.99619 

.oo38o 

63 

53087 

1936 

.96353 

.03646 

14 

98390 

379 

.99614 

.00385 

64 

5ii5i 

2014 

.96062 

.03937 

i5 

98011 

396 

.99595 

.00404 

65 

49 '37 

^080 

.95766 

.04233 

i6 

97615 

426 

.99563 

.00436 

66 

47057 

2i38 

.95456 

.04543 

n 

97189 

469 

.99517 

.00482 

67 

•  44919 

2186 

.95133 

.04866 

i8 

96720 

.525 

.99457 

.00542 

68 

42733 

2224 

.94795 

.o52o4 

19 

96195 

58 1 

.99396 

.oo6o3 

69 

4o5o9 

2268 

.94401 

.05598 

20 

95614 

621 

.99350 

.00649 

70 

38241 

233i 

.93904 

.06095 

21 

94993 

645 

.99321 

.00679 

71 

35910 

2401 

.93313 

.06686 

22 

94348 

653 

.99307 

.00692 

72 

33509 

2469 

.92631 

.07368 

23 

93695 

65i 

.9930  5 

.00694 

73 

3 1 040 

253i 

.91846 

.08154 

24 

93044 

647 

.99304 

.00695 

74 

28509 

2567 

.90995 

.09004 

25 

92397 

647 

.99299 

.00700 

75 

25942 

2542 

.90201 

.09798 

26 

91750 

65 1 

.99290 

.00709 

76 

23400 

2476 

.89418 

.io58i 

^^ 

91099 

668 

.99266 

.00733 

77 

20924 

2369 

.88678 

.11321 

90431 

686 

.99241 

.00758 

78 

18555 

2247 

.87890 

.12109 

29 

89745 

703 

.99216 

.00783 

79 

i63o8 

21 10 

.87061 

.12938 

3o 

89042 

718 

.99193 

.00806 

80 

14198 

1969 

.86i3i 

.13868 

3i 

88324 

726 

.99178 

.00821 

81 

12229 

1823 

.85092 
.83932 
.82573 

.14907 

32 

87598 

733 

.99163 

.00836 

82 

10406 

1672 

.16067 

33 

86865 

743 

.99144 

.00855 

83 

8734 

l522 

.17426 

34 

86122 

754 

.99124 

.00875 

84 

7212 

i36o 

.81142 

.18857 

35 

85368 

768 

.99100 

.00899 
.00932 

85 

5852 

1186 

,79733 

.20266 

36 

84600 

789 

.99067 

86 

4666 

1014 

.78268 

.21731 

ll 

838ii 

811 

.99032 

.00967 

87 

3652 

849 

.76752 

.23247 

38 

83000 

83o 

.99000 

.01000 

88 

2803 

689 

.75419 

.24580 

39 

82170 

844 

.98972 

.01027 

89 

2114 

548 

•74077 

.25922 

40 

8i326 

854 

.98949 

.oio5o 

90 

1 566 

435 

.72222 

.27777 

41 

80472 

860 

.98931 

.01068 

91 

ii3i 

336 

.70291 

.29708 

42 

79612 

869 

.98908 

.01091 

92 

It 

247 

.68930 

.3 1 069 

43 

78743 

888 

.98872 

.01127 

93 

181 

.66970 
.643o5 

.33029 

44 

77855 

913 

.98827 

.01172 

94 

36-1 

i3i 

.35694 

45 

76942 

948 

198698 

.01232 

95 

236 

86 

.63559 

.36440 

46 

75994 

989 

.oi3oi 

96 

i5o 

56 

.62666 

.37333 

H 

75oo5 

1029 

.98628 

.01371 

97 

fo 

44 

.53191 

.46808 

48 

73970 

1067 

.98557 

.01442 

98 

33 

.34000 

.66000 

49 

72909 

1102 

.98488 

.oi5i  1 

99 

17 

II 

Vs 

% 

DO 

71807 

ii33 

.98422 

.01577 

100 

6 

4 

H 

% 

5i 

70674 

1167 

.98348 

.oi65i 

101 

2 

2 

52 

69507 

1204 

.98267 

.01732 

102 

0 

.... 

53 
54 

683o3 
67052 

I25l 

i3o4 

.98168 
.98055 

.oi83i 
.01944 

Note.  The  abo\ 

e  table  is 

that  of 

55 

65748 

1358 

.97934 
.97804 

.02065 

the 

English  Institu 

te  of  Act 

iiaries, 

56 

57 
58 
59 

64390 

1414 

.02195 

pre 

pared  between  18 

62andl86S 

,  from 

62976 
6i5o5 
59974 

147 1 
i53i 
1601 

.97664 
.97510 
.97330 

.02335 
.02489 
.02669 

the 
lea 

continued  expei 
ding  life  insuranc 

'ience  of  t 
le  compan 

wenty 
ies. 

PROBABILITIES.  819 

Pkoblem.  To  find  the  probability  that  a  person  of  age  a 
will  live  to  age  y. 

Solution.  We  take  from  the  table  the  number  living  at 
age  y,  and  divide  it  by  the  number  living  at  age  a.  The  quo- 
tient is  the  probabiUty. 

374:,  The  principle  on  which  the  value  of  a  contingent 
payment  is  determined  is  the  following : 

Theorem.  The  value  of  a  -prohdble  payiiieut  is  equal 
to  the  sum  to  he  paid,  multiplied  hy  the  pi^ohahility  that 
it  will  he  paid. 

Proof,  Let  there  be  n  men,  for  each  of  whom  there  is  a 
probability  p  that  he  will  receive  the  sum  s.  Then  by  §  272, 
Th.  I,  pn  of  the  men  will  probably  receive  the  payment,  so  that 
the  total  sum  which  all  will  receive  will  probably  be  /j?k^.  Now, 
before  they  know  who  is  to  get  the  money,  the  value  of  each 
one's  share  is  equal.  Therefore,  to  find  this  value,  we  divide 
the  whole  amount  to  be  received,  namely,  pns,  by  the  number 
of  men,  n.  This  gives  ps  as  the  value  of  each  one's  chance, 
which  proves  the  theorem. 

Note.  In  this  proof  it  is  tacitly  supposed  that  the  pns 
dollars  are  as  valuable  divided  among  the  pn  men  as  divided 
among  all  n  men.  But  this,  though  supposed  in  mathematical 
theory,  is  not  morally  true.  Morally,  the  money  will  do  more 
good  when  divided  among  all  the  men  than  when  divided 
among  a  portion  selected  by  chance.  All  gambling,  whether 
by  lotteries  or  games  of  chance,  is  in  its  total  efiects  upon  the 
pecuniary  interests  of  all  parties  a  source  of  positive  disadvan- 
tage. This  disadvantage  is  treated  mathematically  by  more 
advanced  methods  in  special  treatises. 


EXERCISES. 

.  Find  from  the  table  th 

e  probabilities 

5  that  a 

person 

a. 

Aged 

30 

will  live   to 

70. 

I, 

i6 

30 

6i                     (i 

80. 

c. 

ii 

50 

(i                     i( 

60. 

d. 

(C 

60 

((                     ii 

70.     . 

320  PROBABILITIES. 


e. 

Aged 

.   70   will  live  to   80. 

/. 

a 

80     "          "       90. 

9- 

66 

90      "          "       95. 

h. 

a 

95      "          "     100. 

2.  What  age  is  that  at  which  it  is  an  even  chance  whether 
a  person  aged  40  will  be  living  or  dead  ? 

3.  Show  that  the  probability  that  a  person  aged  30  will  live 
to  70  is  equal  to  the  product  of  the  probability  that  he  will  live 
to  60  multiplied  by  the  probability  that  a  man  aged  60  will 
live  to  70.     (Apply  the  theorem  of  §  269.) 

4.  What  premium  ought  a  man  of  65  to  pay  for  insuring 
his  life  for  $7000  for  1  year  ? 

5.  Ten  young  men  of  25  form  a  club.  What  is  the  proba- 
bility that  it  will  be  unbroken  by  death  for  ten  years  ? 

6.  The  probability  that  a  planing  mill  will  burn   down 

within  any  one  year  is  -•     What  ought  an  insurance  company 
o 

to  charge  to  insure  it  to  the  amount  of  $3000  for  1  year,  for 
2  years,  for  3  years,  and  for  4  years,  respectively  ? 

7.  If  the  probability  that  a  house  will  burn  down  in  any 

one  year  is  ^,  what  ought  to  be  the  premium  for  insuring  it 

for  s  years  to  the  amount  of  a  dollars  ? 

Note.  In  cases  like  the  last  two,  it  is  assumed  that  only  one  loss 
will  be  paid  for. 

8.  What  is  the  probability  that  if  a  man  aged  25  marry  a 
wife  of  20,  they  will  live  to  celebrate  their  golden  wedding? 

9.  A  company  insures  the  joint  lives  of  a  husband  aged  70 
and  a  wife  aged  50  for  $5000  for  5  years,  the  stipulation  being 
that  if  either  of  them  die  within  that  time  the  other  shall  be 
paid  the  money.  What  ought  to  be  the  premium,  no  allow- 
ance being  made  for  interest  ? 

10.  A  man  aged  50  insures  the  life  of  his  wife,  aged  35,  for 

$10,000  for  20  years,  with  the  promise  that  the  money  is  not 

to  be  paid  unless  he  himself  lives  to  the  age  of  70.     What 

ought  to  be  the  premium  ? 

Note.  In  computations  relating  to  the  management  of  life  insurance, 
it  is  always  necessary  to  allow  compound  interest  on  all  paym(;nts.  But 
the  above  exercises  are  intended  only  to  illustrate  the  application  of  the 
theory  of  probabilities  to  the  subject,  and  therefore  no  allowance  for  in- 
terest is  expected  to  bu  made  in  the  answers. 


BOOK    XI. 

OF  SERIES  AND  THE  DOCTRINE   OF 
LIMITS. 


CHAPTER    I. 

NATU  RE    OF    A     SERIES. 

275.  Def.  A  Series  is  a  succession  of  terms  follow- 
ing each  other  according  to  some  general  law. 

Examples.  An  arithmetical  progression  is  a  series  deter- 
mined by  the  law  that  each  term  shall  be  greater  than  the 
preceding  one  by  the  same  amount. 

A  geometrical  progression  is  a  series  subject  to  the  law 
that  the  ratio  of  every  two  consecutive  terms  is  the  same. 

These  two  progressions  are  the  simplest  form  of  series. 

A  series  may  terminate  at  some  term,  or  it  may  continue 
indefinitely. 

Def,  A  series  which  continues  indefinitely  is  called 
an  Infinite  Series. 

Def.  The  Sum  of  a  series  is  the  algebraic  sum  of 
all  its  terms.  Hence  the  sum  of  an  infinite  series  will 
consist  of  the  sum  of  an  infinite  number  of  terms. 

276.  The  law  of  a  series  is  generally  such  that  the  n^^ 
term  may  be  expressed  as  a  function  of  n. 

For  example,  in  the  series 

1.1.1.1 

2+3+4  +  5  +  '*"- 

the  n^^  term  is  r« 

/J  +  1 

21 


322  SERIES. 


In  the  series      --^  +  ^-^  +-3. "4  +  ^^^'^ 


the  'nP^^  term  is 


n  {n  +  1) 

_Z>(g/;  The  expression  for  the  n^^  term  of  a  series  as 
a  function  of  7^  is  called  the  General  Term  of  the 
series. 

EXERCISES. 

Express  the  n^^  term  of  each  of  the  following  series  : 

'•     3^+4^  +  5^  +  ^*"- 

2.     1-3 +  3-4 +  5-6  +  etc. 

3-  1  +  ra  +  r?.  3  + '^^^^ 

IX  ft.  i)u  ^ 

Write  four  terms  of  each  of  the  series  having  the  following- 
general  terms : 

4^^2  _  X 

5.  The  n^^  term  to  be  — -r, 7- 

^  47i^  +  1 

6.  The  i^^'  term  to  be  «:  (i.  -j-  1)  (^  +  2)  ^. 

(^  +  3)  (/^  +  4)  2:^+1 


7.  The  (^  +  ly^  term  to  be 

8.  The  {n  —  ly^  term  to  be 


{n  +  5)  (^  +  6) 


1-2 ^ 

1 

277.  The  most  common  nse  of  a  series  is  to  enable  us  to  j 

compute,  by  approximation,  the  values  of  expressions  which  it 

is  difficult  or  impossible  to  compute   directly.     Suppose,  for 

1  -{-  X 
example,  that  we  have  to  compute  the  value  of when  x 

is  a  small  fraction,  say  — ,  and  to  have  the  result  accurate  to 

eight  decimals.     We  shall  see  hereafter  that  when  x  is  less  than 
1,  we  have 


CONVERGENCE   OF  8EIUES.  82c 

=  1  -\- 2x  -}- 2x^  -i-  2x^  +  etc.,  ad  infaiitum. 


1—x 

50 


Suppose  X  z=  —  =z  ,02.    We  compute  this  series  thus: 

1 


2  X  .02  =: 

.04 

Multiplying  by  .02, 

.0008 

a                 (e 

.000016 

i(                       <£ 

.00000032 

^             1.02 
Sum  =      -^  = 

1.04081632 

.098 
which  IS  much  more  expeditious  than  dividing  1  02  by  .98. 

It  will  be  seen  that  every  term  we  add  makes  the  quotient 
accurate  to  one  or  two  more  decimals,  so  that  there  is  no  limit 
to  the  precision  which  may  be  attained  by  the  use  of  the  series. 

If,  however,  x  had  been  greater  than  unity,  the  series  would 
give  no  result,  because  the  terms  2x,  2x%  2x%  would  have  gone 
on  increasing  indefinitely,  whereas  the  true  value  of  the  frac- 

1  -{-  X 
tion would  have  been  negative. 

This  example  illustrates  the  following  two  cases  of  series : 

I.  There  may  he  a  certain  limit  to  ivhieh  the  sum  of 
the  series  shall  approach,  as  we  increase  the  numher  of 
terms,  hut  which  it  can  never  reach,  how  great  soever  the 
number  of  terms  added. 

For  example,  the  series  we  have  just  tried, 

2         2         2         2 

^  +  50  +  50-^  +  50-3  +  505  +  "*''-' 

1  02 
approaches  the  limit  tt-^t.,  but  never  absolutely  reaches  it. 

II.  As  we  increase  the  numher  of  terms,  the  sum 
may  increase  without  limit,  or  may  vihrate  hach  and 
forth  in  consequence  of  some  terms  heing  positive  and 
others  negative. 

These  two  classes  of  series  are  distinguished  as  convergent 
and  divergent. 


324  SERIES. 

Def,  A  Convergent  Series  is  one  of  whicli  the  sum 
approaches  a  limit  as  the  number  of  terms  is  increased. 

Refer  to  §  213  for  an  example  of  infinite  series  in  geometrical  pro- 
gressions whicli  have  limits. 

Def.  A  Divergent  Series  is  one  of  which  the  sum 
does  not  approach  a  limit. 

Examples.  The  series  1-f  2-f  3  +  4-f  etc.,  ad  iJifaiihirn, 
IS  divergent,  because  there  is  no  limit  to  the  sum  of  its  terms. 

The  series  1  —  1+  1  —  1  +  1—  etc.,  is  divergent,  because 
its  sum  continually  fluctuates  between  +1  and  0. 

Rem.  When  we  consider  only  a  limited  number  of  terms, 
the  question  of  convergence  or  divergence  is  not  important. 
But  when  the  sum  of  the  whole  series  to  infinity  is  to  be  con- 
sidered, only  convergent  series  can  be  used. 

Notation  of  Stims. 

378.  The  sum  of  a  series  of  terms  represented  by 
common  symbols  may  be  expressed  by  the  symbol  2, 
followed  by  one  of  the  terms. 

Example.     The  expression 

^a 
means  "the  sum  of  several  terms,  each  represented  by  «." 

When  it  is  necessary  to  distinguish  the  different 
terms,  different  accents  or  indices  are  affixed  to  them, 
and  represented  by  some  common  symbol. 

Example.    The  expression 

means  the  sum  of  several  terms  represented  by  the  symbol  a 
with  indices  attached  ;  that  is,  the  sum  of  several  of  the  quan- 
tities a^y  ^3,  «3,  a^,  etc. 

When  the  particular  indices  included  in  the  summa- 
tion are  to  be  expressed,  the  greatest  and  least  of  them 
are  written  above  and  below  the  symbol  2. 


SIGJS'   OF  SUMMATION,  325 

Examples.     The  expression 

i=15 
i=5 

means :  "  Sum  of  all  the  symbols  ai  formed  by  giving  i  all  in- 
tegral values  from  i  =  5  to  i=  15."     That  is, 

e=15 
lai=zas  +  ^6  -r  ^7  +  <^8  +  «9  +  ^10 +  «^ii  + ^12 +  ^13+^14  +  ^15' 

i=5 
i=5 

l^irn  means  0  +  m  +-  2m  +-  3/>2  +  4//z  +-  5/^.. 

i=0 

's  (i,y)  means  (1,/)  +  (2,7)  +  (3,/)  +  (4,^). 
%-\ 

'~Hhj)  =  {t,  2)  +  {h  3)  +  (^,  4)  +  (^,  5)  +  {i,  6). 
''i?^!  —  l!  +  2!+-3!  +  4!  =  1  +  2  +  6  +  24  =  33. 
'~2^  =  7  +  8  +  9  +-  10  +■  11  =  45. 
'  2^2  =  22  +  32  +  42  +  5^  =  54. 

i=2 

EXERCISES. 

Write  out  the  following  summations,  and  compute  theii 

values  when  they  are  purely  numerical : 

i=7  n=Q  n=6 

I.       ^j\  2.       ^nin-^l),       3.       2^(^  +  1). 

i=8  n=7  «=6 

4.     2wi.         5.     2n>fc.  6.     l:(7^+l)(y— 1). 

i=4  ?i=5  «=5  ^ 2 

7.       ^imi,  8.       2/^%2^  9.       2  — — -• 

i=2  w=2  71=0  ^  +  1 

Express  the  following  sums  by  the  sign  2 : 

10.     h^+lH+h^+h^+h^.  lu     13  +  23  +  33  +  48. 

12.     1.2  +  2.3  +  3.4  +  4.5.       13.     I  +  I  +  I  +  J  +  I 


326  8EUIES. 

CHAPTER     II. 

DEVELOPMENT    IN    POWERS    OF    A    VARIABLE. 

379.  Among  the  most  common  series  employed  in  math- 
ematics are  those  of  which  the  terms  are  multiplied  by  the 
successive  powers  of  some  one  quantity. 

An  example  of  such  a  series  is 

1 -\- ^z -\- dz^  -{-  4^3  +  5;2^  +  etc., 

in  which  each  coefficient  is  greater  by  unity  than  the  power  of 
z  which  it  multiplies. 

A  geometrical  progression,  it  will  be  remarked,  is  a  series 
of  this  kind,  in  which  the  terms  contain  the  successive  powers 
of  the  common  ratio. 

The  general  form  of  such  a  series  is  j 

in  which  the  successive  coefficients  a^,  a^,  a^,  etc.,  are  formed 
according  to  some  law,  but  do  not  contain  z. 

Such  a  series  as  this  is  said  to  proceed  according  to  the 
ascending  powers  of  the  variable  z, 

Eem.  The  sum  of  a  series  is  often  equal  to  some  algebraic 
expression  containing  the  variable.  Conversely,  we  may  find  a 
series  the  sum  of  all  the  terms  of  which  shall  be  equal  to  a  . 
given  expression. 

Def.    A  series  equal  to  a  given  expression  is  call< 
the  Development  of  that  expression. 


e^ 


To  Develop  an  expression  means  to  find  a  series 
the  sum  of  all  the  terms  of  which  are  equal  to  the  ex- 
pression. 

The  most  extensively  used  method  of  development  is  that 
of  indeterminate  coefficients. 


INDETEBMINATE   COEFFICIENTS.  327 

Method  of  Indeterminate  Coefficients. 

380.  The  method  of  indeterminate  coefficients  is  based 
upon  the  following  principles  : 

Let  us  have  two  equal  expressions,  each  containing  a  varia- 
ble z,  and  one  or  both  containing  also  certain  indeterminate 
quantities^  that  is,  quantities  introduced  hypothetically,  and  not 
given  by  the  original  problem,  the  values  of  which  are  to  be 
subsequently  assigned  so  as  to  fulfil  a  certain  condition. 

The  condition  to  be  fulfilled  by  the  values  of  the  inde- 
terminate quantities  is  that  the  two  expressions  containing  z 
and  these  quantities  shall  be  made  identically  equal. 

Then,  because  the  equations  are  to  be  identically  equal,  wo 
can  assign  any  values  we  please  to  z,  and  thus  form  as  many 
equations  as  we  please  between  the  indeterminate  quantities. 

If  these  equations  can  be  all  satisfied  by  one  set  of  values  of 
these  quantities,  then  by  assigning  these  values  to  them  in  the 
original  equation,  the  latter  will  be  an  identical  one,  as  required. 

The  student  should  trace  the  above  general  method  in  the  following 
examples  of  its  application. 

281.  Theorem  I.  //  a  series  proceeding  according 
to  the  ascending  -powers  of  a  quantity  is  equal  to  zero  for 
all  values  of  that  quantity,  the  coefficient  of  each  sepa- 
rate term  must  he  zero. 

Proof,  Let  the  several  coefficients  be  a^,  a^,  ^g,  etc.,  and 
z  the  quantity,  so  that  the  series,  put  equal  to  zero,  is 

a^  -f-  a^z  +  (^2^2  -f  a^z^  +  etc.  =  0. 

Because  the  equation  is  true  for  all  values  of  z,  it  must  be 
true  when  z=zO.     Putting  z  =  0,  it  becomes 

a^  =  0. 
Dropping  a^,  the  equation  becomes 

a^z  +  a2Z^  +  a^z^  -f-  etc.  =  0. 
Dividing  by  z,     a^  +  a2Z  +  a^z^  +  etc.  =  0. 
From  this  we  derive,  by  a  repetition  of  the  same  reasoning, 
a^  =  0. 


328  SERIES. 

Continuing  the  process,  we  find 

«2  =  0,    ^3  =  0,     etc.,  indefinitely. 

Theoeem  II.  //  two  series  proeeeding  hy  ascending 
powers  of  a  quantity  are  equal  for  all  values  of  that 
quantity,  the  coefficients  of  the  equal  powers  must  he 
equal. 

Proof,     Let  the  two  equal  series  be 

aQ-\-a^z-\-a^z^-\-eiG,  =  Z>o+^i^+^2^^  +  ^^c.  {a) 

Transposing  the  second  member  to  the  left-hand  side  and 
collecting  the  equal  powers  of  z,  the  equation  becomes 

^0  —  ^0  +  (^1  —  ^i)  ^  +  (^2  —  ^3)  ^^  +  etc.  =  0. 
Since  this  equation  is  to  be  satisfied  for  all  values  of  z,  the 
coefficients  of  the  separate  powers  of  z  must  all  be  zero. 

Hence, 

^0  ■""  ^0  =  ^9        a^  —b^  =  0,        a^  —  lc^  =  0,       etc. 
or  a^  =:  h^,  a^  =  b^,  a^  =  h^,     etc. 

ExEECiSE.  Let  the  student  demonstrate  these  last  equa- 
tions independently  from  (a),  by  supposing  2;  =  0,  then  sub- 
tracting from  both  sides  of  {a)  the  quantities  found  to  be  equal ; 
then  dividing  by  z  ;  then  supposing  ^  =  0,  etc. 

Rem.  The  hypothesis  that  {a)  is  satisfied  for  all  values  of 
z  is  equivalent  to  the  supposition  that  it  is  an  identical  equa- 
tion. In  general,  when  we  find  different  expressions  for  the 
same  functions  of  a  variable  quantity,  these  expressions  ought 
to  be  identically  equal,  because  they  are  expected  to  be  true 
for  all  values  of  the  variable. 

Theorem  III.  A  function  of  a  variable  can  only  he 
developed  in  a  single  way  in  ascending  powers  of  the 
variahle. 

For  if  we  should  have 

Fz  =  A^-\-  A^z  +  A^z^  +  A^z^  +  etc., 
and  also      Fz  =  B^  -{-  B^z  +  B^z^  +  B^z^  +  etc., 


INDETERMINATE   COEFFICIENTS.  329 

these  two  series,  being  each  identically  equal  to  Fz,  must  be 
identically  equal  to  each  other.  But,  by  Th.  II,  this  cannot  be 
the  case  unless  we  have 

Aq  =z  Bq,    J-i  =  B^,     A^  =  B^,    etc. 

The  coefficients  being  equal,  the  two  series  are  really  one 
and  the  same. 

383.  Expansion  ly  Indeterminate  Coefficients,  The  above 
principle  is  applied  to  the  development  of  functions  in  powers 
of  the  variable.  The  method  of  doing  this  will  be  best  seen 
by  an  example. 

1.  Develop in  powers  of  x, 

JL   "Y"  X 

Let  us  call  the  coefficients  of  the  powers  of  x  a^,  a^,  etc. 
The  series  will  be  known  as  soon  as  these  coefficients  are 
known.    Let  us  then  suppose 

— - —  =  a^  +  a^x  +  a^oG^  +  a^o(^  +  etc. 

J-  -J—  X 

Here  we  remark  that,  so  far  as  we  have  shown,  this  equa- 
tion is  purely  hypothetical.  We  have  not  proved  that  any 
such  equation  is  possible,  and  the  question  whether  it  is  possi- 
ble must  remain  open  for  the  present.  We  must  find  whether 
we  can  assign  such  values  to  the  indeterminate  coefficients,  a^, 
a^,  a 2,  etc.,  that  the  equation  shall  be  identically  true. 

Assuming  the  equation  to  be  true,  we  multiply  both  sides 
by  1  -^  X.     It  then  becomes 

I  =  a^  -^  {Gq  +  a^)  X  -\-  (a^  -}-  a^)  x^  +  etc. ; 
or  transposing  1, 
0  =  t?o  -r-  1  +  {aQ-\-a^)x  +  {a^-\-a2)x^  +  {a2+a^)a^  -\-  etc. 

By  Theorem  I,  the  coefficients  must  be  identically  zero. 
Hence, 

a^  —  1    =0,    which  gives 
a^-\.a,  =  0, 
a^  -I-  a,  =  0, 
a^^a^  =  0, 
etc. 


«o 

= 

1; 

^1 

=: 

-«« 

:= 

— 

1; 

«2 

:= 

—  «i 

= 

1; 

«3 

etc. 

= 

— 

1; 

330  SERIES. 

Substituting  these  values  of  the  coefficients  in  the  original 
equation,  it  becomes 

=  1  —  x-\-x^^a^-\-o(^---  etc. 

1  +  ^ 

This  same  method  can  be  applied  to  the  development  of 
any  rational  fraction  of  which  the  terms  are  entire  functions 
of  some  one  quantity.     Let  us,  for  instance,  suppose 

m  -\-  nx  -\-  im^  o  i  ^       . 

Multiplying  by  the  denominator  of  the  fraction,  this  equa- 
tion gives 

a  +  1)X  ^^  mA^  +  (nA^-^mA^  x  +  {^^pA^-^nA^  +mA^)x^ 

+  (pA  1  +  /^^  2  +  mA  s)x^  -{■  etc. 

We  now  see  that  when  i  >  1,  the  coefficient  of  x^  in  this 
equation  is  mAi  +  nAi_t  +  pAi-2' 

Equating  the  coefficients  of  like  powers  of  x, 

mAr.  =  a,  whence     Ar.  =  ~; 
mA^  +  tiAq  =  h,  " 

mAc^  +  nA^  +P^o  =  ^^  " 

VI A ^  4-  nA^  +pA^  =  0,  ^^ 

We  have  from  the  general  coefficient  above  written,  when 

Ai= Ai-t  —  ^Ai-2. 

m  m 

That  is,  each  coefficient  after  the  second  is  the  same 
linear  function  of  the  two  coefficients  next  preceding. 

Such  a  series  is  called  a  Recurring  Series. 

EXERCISES. 

Develop  by  indeterminate  coefficients : 

1  _1 

^'     1  —  x  ^'     1  —  2^' 


^1 

— 

m 

— 

""a 

? 

^2 

= 

— 

P. 
m 

^0- 

m    ^^ 

^3 

= 

— 

L 
m 

^1- 

m    ^ 

UNDETERMINED    MULTIPLIERS,  331 

1—X  1  -\-  X 


1  +  X 

1  +  x 

4. 
6. 

8. 

1-x 
1-x 

1  4_  2a:  -f  3^:2 
1  —  2^  +  S^^ 

l  —  2x-\-x^ 
1-x 

1  +  2:z;  +  dx^ 

I  ^x  —  x^ 

383.  The  development  of  a  rational  fraction  may  also  be 
effected  by  division,  after  the  manner  of  §§  96,  97,  the  opera- 
tion being  carried  forward  to  any  extent. 

Example.     Develop  j-^ — 

-L  —  X 

1+X  \l  —  x  

1—X 


2x 

2x  —  2x^ 


1  +  ^^  +  2x^  +  2x^  +  etc. 


2a;2  +  0 
2x^  —  2x^ 

2x^,  etc. 

EXERCISES. 

Develop  by  division  the  expressions : 

1  —  2^  1  -\-x 


I. 


1  -\-  X  1  —X  -\-  x^ 


384.  Elimination  hy  Undetermined  Multipliers,  There  is 
an  application  of  the  method  of  undetermined  coefficients  to 
the  problem  of  eliminating  unknown  quantities,  which  merits 
special  attention  on  account  of  its  instructiveness.  Let  any 
system  of.  simultaneous  equations  between  three  unknown 
quantities  be 

ax  -\-    ly  -\-    cz  =^  Ji,  (1) 

a'x  -h  h'y  +  cz  =  7i',  (2) 

a"x  +  b^'y  +  c'^z  =  It",  (3) 

Can  we  find  two  such  factors  that,  if  we  multiply  two  of 
the  equations  by  them,  and  add  the  results  to  the  third,  two  of 
the  three  unknown  quantities  shall  be  eliminated  ? 


332  SERIES. 

This  question  is  answered  in  the  following  way : 

If  there  be  such  factors,  let  us  call  them  m  and  n.     If  we 

multiply  the  first  equation  by  m,  the  second  by  7i,  and  add  the 

product  to  the  third  equation,  we  shall  have 

{am  +  a'ji  +  a")  x  ) 
+  {b7n  +  b'n  +  b")  y  ^  =  hn  +  h'71  +  h'\  (b) 

+  {cm  4-  en  +  c")  z  ) 

In  order  that  the  quantities  ^  and  ^  may  disappear  from 
this  equation,  we  must  have 

hn  +  b'n  +  b"  =  0, 
cm  +  c'n  -j-  c"  =  0. 

Since  we  have  these  two  equations  between  the  quantities 
m  and  n,  we  can  determine  their  values. 
Solving  the  equations,  we  find : 

h'c"  -  b"c' 


m 


n  = 


Id  -Vc  ' 

Vc  -  he" 
be'  -  b'c  ' 


These  are  the  required  values  of  the  multipliers.  Substi- 
tuting them  in  the  equation  {b),  we  find  that  the  coefficients 
of  2/  and  2;  vanish,  and  that  the  equation  becomes 


[' 


a{b'e"  —  b"e')  -i-a'  {b"e-  be") 

bd^  b'c  "^  "^ 


k  {b'c"  -  b"e')  +  It'  {b"e  -  be")  ^  ^„ 


Clearing  of  denominators  and  dividing  by  the  coefficient  of 
X,  we  find 

__  h  {b'c"  -  b"e')  +  h'  {b"c  -  be")  +  li"  {be'  -  b'e) 
^  -  a  {b'e"  -  b"c')  +  a'  {b"c  -  be")  +  a"  {be'  -  b'e) 


EXERCISES. 


I.  Find  the  values  of  y  and  z  by  the  above  process  for 
finding  x. 


MULTIPLICATION   OF  SERIES.  333 

For  this  purpose  we  may  begin  with  the  equation  (&)  and  find  values 
of  m  and  n  such  that  the  coefficients  of  x  and  z  in  (p)  shall  vanish.  These 
values  will  be  different  from  those  given  in  {c).  By  substituting  them  in 
(6),  X  and  z  will  be  eliminated,  and  we  shall  obtain  the  value  ofy. 

We  then  find  a  third  set  of  values  of  m  and  n^  such  that  the  coeffi- 
cients of  X  and  y  shall  vanish,  and  thus  obtain  the  value  of  z. 

2.  Solve  by  the  method  of  indeterminate  multipliers  the 
exercise  3  of  §  140. 

Multiplication  of  Two  Infinite  Series. 

284a.  Problem.  To  express  the  product  of  the  two 
series 

and  Iq  +  d^x  +  Ic^x^  +  h^x^  +  etc. 

The  method  is  similar  to  that  by  which  the  square  of  an 
entire  function  is  formed  (§  173,  2). 

We  readily  find  the  first  two  terms  of  the  product  to  be 

The  combinations  which  produce  terms  in  x'^  are 

a^b^x^  +  a^l^x^  -f  a^h^x^. 
Those  which  produce  terms  in  x^  are 

^0^3^  +  ct'J)^x^  +  ^2^1^  +  a^l^x^. 

In  general,  to  find  the  terms  in  x'^  we  begin  by  multiplying 
Gq  into  the  term  InX^  of  the  lower  series,  and  then  multiplying 
each  succeeding  of  the  first  series  by  each  preceding  term  of 
the  second,  until  we  end  with  anb^x'^.     Hence,  if  we  suppose 

Product  =  Aq  -\-  A^x  -\-  A^x^  -]-.,,  ,  +  AnX"^  +  etc., 

we  shall  have,  for  all  values  of  n, 

An  =  a^bn  +  a^bn-i  +  a^bn-%  +  ....+  anb^. 

By  giving /I all  integral  values,  we  shall  form  as  many  values 
as  we  choose  of  J^,  and  so  as  many  terms  as  we  choose  of  the 
series. 


334  SERIES. 

EXERCISES. 

1.  Form  the  product  of  the  two  series: 

^        x^       oc^       ofi        , 

^  "~  2!  +  4!  ~  6l  +  ^^""'^ 

/y»3  /y^  '^-1 

tAj  tlU  >Aj  . 

^  ~~  3~!  "^  5l  "~  7l  "^ 

2.  Form  the  square  of  each  of  these  series. 

3.  Can  you,  by  adding  the  squares  together,  show  that  their 
sum  is  equal  to  unity,  whatever  be  the  value  of  ir? 

To  effect  tins,  multiply  each  coefficient  of  x'^  in  the  sum  of  the  squares 
by  n\ ,  substitute  for  each  term  its  value  CT  given  in  §  257,  and  apply 
§  262,  Th.  II. 

385,  Series  proceeding  according  to  the  Powers  of  Two 
VariaUes.    Such  a  series  is  of  the  form 

in  which  the  products  of  all  powers  of  x  and  y  are  combined. 
By  collecting  the  coefScients  of  each  power  of  x,  the  series  will 
become 

+  (^0  +  (^\y  +  ^2^^  +  <^3^^  +  — )^^ 

+  etc.^     etc.,      etc.,      etc. 

Hence,  the  series  is  one  proceeding  according  to  the  powers 
of  one  variable,  in  which  the  coefficients  are  themselves  series, 
proceeding  according  to  the  ascending  powers  of  another 
variable. 

Let  us  have  the  identically  equal  series  proceeding  accord- 
ing to  the  ascending  powers  of  the  same  variables, 

+  etc.,      etc.,       etc. 

Since  these  series  are  to  be  equal  for  all  values  of  x,  the 
coefficients  of  like  powers  of  x  must  be  equal.     Hence, 


8EBIE8.  335 

«o  +  f^iV  +  «s^^  +  etc.  =  ^0  +  ^iV  +  -^32/^  +  e^c- 
^0  +  ^1^  +  ^3^^  4-  etc.  =  ^0  +  ^1^  +  B^y^  +  etc. 
etc.  etc. 

Again,  since  these  series  are  to  be  equal  for  all  values  of  y, 
we  must  have 


«o    —   ^0^ 

«i  ==  A^, 

(^2    =   ^3^ 

etc, 

d,  =  B,, 

h  =  ^1, 

^,  =  B,, 

etc. 

etc. 

etc. 

etc. 

Hence,  in  order  that  two  series  proceeding  according 
to  the  ascending  powers  of  two  variables  may  he  identi- 
cally equal,  the  coefficients  of  every  liJce  product  of  the 
powers  must  he  equal. 


336  SERIES. 

CHAPTER     III. 
SUMMATION      OF     SERIES. 


Of  Figiirate  Numbers. 

386.    The  numbers  in  the  following  columns  are  formed 
according  to  these  rules  : 

1.  The  first  column  is  composed  of  the  natural  numbers, 
1,  2,  3,  etc. 

2.  In  every  succeeding  column  each  number  is  the  sum  of 
all  the  numbers  above  it  in  the  column  next  preceding. 

Thus,  in  the  second  column,  the  successive  numbers  are : 

1,     1  +  2  =  3,     1  +  2  +  3  =  6,    1  +  2  +  3  +  4  =  10,  etc. 
In  the  third  column  we  have  j 

1,     1  +  3=4,    1  +  3  +  6  =  10,     etc.  ' 

1 


(^) 


1 

3 

3 

1 

1 

3 

4 

1 

G 

5 

1 

4 

10 

6 

10 

15 

7 

5 

15 

20 

35 

21 

6 

21 

35 

7 

etc. 

etc. 

etc. 

It  is  evident  from  the  mode  of  formation  that  each  number 

is  the  difference  of  the  two  numbers  « 

next  above  and  below  it  in  the  col-  ^      ^ 

umn  next  following.  #      •      • 

The  numbers  1,  3,  6,  10,  etc.,  in  •      •      •      • 

the  second  column  are  called  trian-  •      •      •      •      • 

gular  numbers,  because  they  repre-  iv'^  1+2+3+4+5. 


SERIES,  337 

sent  numbers  of  points  which  can  be  regularly  arranged  over 
triangular  surfaces. 

The  numbers  1,  4,  10,  etc.,  in  the  third  columns  are  called 
pyramidal  numbers,  because  each  one  is  composed  of  a  sum 
of  triangular  numbers,  which  being  arranged  in  layers  over 
each  other,  will  form  a  triangular  pyramid. 

All  the  numbers  of  the  scheme  are  called  figurate  num- 
bers. 

The  numbers  in  the  i^^  column  are  called  figurate  numbers 
of  the  i^^^  order. 

387.  If  we  suppose  a  column  of  I's  to  the  left  of  the  first 
column,  and  take  each  line  of  numbers  from  left  to  right  in- 
cUned  upward,  we  shall  have  the  successive  lines  1, 1 ;  1,  2, 1 ; 
1,  3,3,  1,  etc.  These  numbers  are  formed  by  addition  in  the 
same  way  as  the  binomial  coefficients  in  §  171,  2.  We  may 
therefore  conclude  that  all  the  numbers  obtained  by  the  pre- 
ceding process  are  binomial  coefficients,  or  combinatory  expres- 
sions.   This  we  shall  now  prove. 

Theorem.  Tlie  n^^  nuinher  in  the  i^^  column  is  equal 
to  CT'~'  or  to 

y^(7^4-l)^^  +  2) (^  +  /  -  1) 

1.2-3....  I  *  ^^ 

Proof.  Because  the  com.binations  of  1  in  any  number  arc 
equal  to  that  number,  we  have,  when  i  —  1, 

n^^  number  in  1st  column  =:  7^  =  Ci, 

which  agrees  with  the  theorem. 

When  i  =  2,  we  have,  by  the  law  of  formation  of  the 
numbers, 

n^^  number  in  2d  column  =  6'i  +  Ci  +  (7i  +  .  .  .  .  +  Ci, 
which,  by  equation  (a)  (§260,  3),  is  equal  to  Cg^  . 

Therefore  the  successive  numbers  in  the  second  column, 
found  by  supposing  7^  =  1,  7^  =  2,  etc.,  are 

Cl  G%,  G%,....  (fl'\ 
33 


338  SERIES. 

Since  the  n*^  number  in  the  third  column  is  equal  to  the 
sum  of  all  above  it  in  the  second,  we  have 

n^^  number  in  3d  column  ^  Cl^-Cl+G\+  CV^  =  C^^ 

which  still  corresponds  to  the  theorem,  because,  ^hen  i  =:  3^ 
n  -\-  i  —  1  z=z  n  -^  2- 

To  prove  that  the  theorem  is  true  as  far  as  we  choose  to 
carry  it,  we  must  show  that  if  it  is  true  for  any  value  of  t,  it  is 
also  true  for  a  value  1  greater.  Let  us  then  suppose  that,  in 
the  r^^  column  the  first  n  numbers  are 

j^r      ryr+1      ryr+2  pT+n—l 

Since  the  n^^  number  in  the  next  column  is  the  sum  of 
these  numbers,  it  will  be  equal  to  ^  I 

which  is  the  expression  given  by  the  theorem  when  we  suppose 
i  =  r  -^  1. 

Now  we  have  proved  the  theorem  true  when  i  =  3;  there- 
•fore  (supposing  r  =  3)  it  is  true  for  i  =  4.    Therefore  (sup- 
posing r  =  4)  it  is  true  for  i  =:  5,  and  so  on  indefinitely. 

If  in  the  general  expression  (1)  we  put  i  =  2,  we  shall 
have  the  values  of  the  triangular  numbers  ;  by  putting  i  =  3,  ^ 
we  shall  have  the  pyramidal  numbers,  etc.    Therefore, 

71  {n  +  1) 


The  n^^  triangular  number 
The  n^  pyramidal  number 


1.2 

Qi  {n  +  1)  {n  +  2) 


1.2-3 

By  supposing  ^  =  1,  2,  3,  4,  etc.,  in  succession,  we  find 
the  succession  of  triangular  numbers  to  be 

1-2      2.3      4.5 

1.2^    1.2'     1.2'    ^^^'' 

and  the  pyramidal  numbers, 

1.2.3      2.3.4      3.4.5 
1.2.3'     1.2-3'    1.2.3'  ' 

which  we  readily  see  correspond  to  the  values  in  the  scheme  (A). 


8EBIE8.  339 

Enumeration  of  Triangular  Piles  of  Shot. 

388.  An  iuteresting  application  of  the  preceding  theory  is 
that  of  finding  the  number  of  cannon-shot  in  a  pile.     There 
are  two  cases  in  which  a  pile  will  con- 
tain a  figurate  n  umber  \ 

I.  Elongated  projectiles,  in  which 
each  rests  on  two  projectiles  below  it. 

II.  Spherical  projectiles,  each  rest- 
ing on  three  below  it,  and  the  whole 
forming  a  pyramid. 

Case  I.    Elongated  Projectiles,    Here 
the  vertex  of  a  pile  of  one  vertical  layer  will  be  formed  of  one 
shot,  the  next  layer  below  of  two,  the  third  of  three,  etc. 
Hence  the  sum  of  n  layers  from  the  vertex  down  will  be  the 
n^^  triangular  number. 

It  is  evident  that  the  number  of  shot  in  the  bottom  row  is 
equal  to  the  number  of  rows.  Hence,  if  7i  be  this  number, 
and  N  the  entire  number  of  shot  in  the  pile,  we  shall  have, 

_  n  {n  +  1) 
2        ' 
If  the  pile  is  incomplete,  in  consequence  of  all  the  layers 
above  a  certain  one  being  absent,  we  first  compute  how  many 
there  would  be  if  the  pile  were  complete,  and  subtract  the 
number  in  that  part  of  the  pile  which  is  absent. 

Example.     The  bottom  layer  has  25  shot,  but  there  are 

only  11  layers  in  all.    How  many  shot  are  there? 

25*26 
K  the  pile  were  complete,  the  number  would  be  — - — 

There  being  14  layers  wanting  from  the  top,  the  total  number 

14*15 
of  shot  wanting  is  —^ —    Hence  the  number  in  the  pile  is 

til 

_  25-26  — 14-15  _  (14  +  11)(15  +  11)  — 14-15 

2  ~  2 

=  ^m^  +  i^  +  n)  ^  ^^^^ 


340  SERIES. 

Note.  This  particular  problem  coald  have  been  solved  more  briefly 
by  considering  the  number  of  shot  in  the  several  layers  as  an  arithmetical 
progression,  but  we  have  preferred  to  apply  a  general  method. 

EXERCISES. 

1.  A  pile  of  cylindrical  shot  has  7i  in  its  bottom  row,  and  r 
rows.     How  many  shot  are  there  ? 

2.  From  a  complete  pile  having  h  layers,  s  layers  are  re- 
moved.    How  many  shot  are  left  ? 

3.  A  pile  has  n  shot  in  its  bottom  row,  and  m  in  its  top 
row.     How  many  rows  and  how  many  shot  are  there? 

4.  A  pile  has  p  rows  and  Tc  shot  in  its  top  row.  How  many 
shot  are  there  ? 

5.  Explain  the  law  of  succession 
of  even  and  odd  numbers  in  the  se- 
ries of  triangular  numbers. 

6.  How  many  balls  are  necessary 
to  fill  a  hexagon,  having  n  balls  in 
each  side  ? 

Note.     In  the  adjoining  figure, 

389.  Ca^se  II.  Pyramid  of  Balls,  If  a  course  of  balls 
be  laid  upon  the  ground  so  as  to  fill  an  equilateral  triangle^ 
having  n  balls  on  each  side,  a  second  course  can  be  laid  upon 
these  having  n  —  \  balls  on  each  side,  and  so  on  until  we 
come  to  a  single  ball  at  the  vertex. 

Commencing  at  the  top,  the  first  course  will  consist  of  1 
ball,  the  next  of  3,  the  third  of  6,  and  so  on  through  the  tri- 
angular numbers.  Because  each  pyramidal  number  is  the 
sum  of  all  the  preceding  triangular  numbers,  the  whole  num- 
ber of  balls  in  the  n  courses  will  be  the  n^  pyramidal  number, 

or 

n{n  +  \)(n  +  ^) 

^  ~"  1.2.3 

EXERCISES. 

I.  How  many  balls  in  a  triangular  pyramid  having  9  balls 
on  each  side  ? 


8ERIE8.  341 

2.  If  from  a  triangular  pyramid  of  n  courses  Jc  courses  be 
removed  from  the  top,  how  many  balls  will  be  left  ? 

3.  How  many  balls  in  the  frustum  of  a  triangular  pyramid 
having  n  balls  on  each  side  of  the  base  and  m  on  each  side  of 
the  upper  course  ? 

Sum  of  the  Similar  Powers  of  an  Arithmetical 
Progression. 

390.   Put        «!,  the  first  term  of  the  progression ; 
di  the  common  difference; 
n,  the  number  of  terms ; 
m,  the  index  of  the  power. 

It  is  required  to  find  an  expression  for  the  sum, 

<  +  («i  +  d)"^  +  (^1  +  ^^)'^  + +  [«^i  +  (^  -  1)  clY", 

which  sum  we  call  Sm- 

Let  us  put,  for  brevity,  a^,  a^,  a^,  a^,  ,  .  ,  .  an  for  the  sev- 
eral terms  of  the  progression.     Then 

«2  =  ^1  +  ^^ 

a^  =:  a^  +  2d  =.^2  +  ^^ 

an  =  a^  +  (^^  —  l)d=z  an-i  +  d. 

Raising  these  equations  to  the  (m  +  iy^  power,  and  adding 
the  equation  an^  =z  an  +  d,  we  have 

am+i  ^  «m+i  -I-  (m  +  1) a^d  +  ^H^+^lHl af-H^  +  etc 
^^+1  =:  a^+i  +  (m  +  1) afd  +  ^^^^^^ af-^d'^  +  etc. 
^m+i-  ^  ^m+i  +  (^  4.  1)  a^d  4-  ^^+^)^  ^m-1^2  4.  etc. 

«m^j  ^  ^m+i  ^{m  +  1)  ay  +  l^i+il^  ^m-i^  +  etc. 

If  we  add  these  equations  together,  and  cancel  the  common 
termSj  a^+i  +  a^+^  +  .  .  .  .  -\-  a^+^,  which  appear  in  both 
members,  we  shall* have 


342  SERIE8. 

.    (^  +  1)  m  (m  -  1)  ,30.  . 

i 172^^ '(^^m-2,  etc. 

From  this  we  obtain,  by  solving  with  respect  to  Sm^ 

C:i-«r'      m  m{m-l) 

which  will  enable  us  to  find  Sin  when  we  knoAv  Si,  S2,  »  *  -  - 
Sm-h  that  is,  to  find  the  snm  of  the  n^^  powers  when  we  know 
the  sum  of  all  the  lower  powers.  It  will  be  noted  that  S^ 
means  the  sum  of  the  arithmetical  series  itself,  as  found  in 
Book  VII,  Chap.  I ;  and  that  Sq  =  n,  because  there  are  71 
terras  and  the  zero  power  of  each  is  1. 

By  §  209,  Prob.  V, 

</ 
To  find  the  sum  of  the  squares,  we  put  m  =z2,  which  gives 

3d  ^^'       3 

391.  The  simplest  application  of  this  expression  is  given 
by  the  problem: 

To  find  the  sum  of  the  squares  of  the  first  n  natural 
ninnbers,  namely, 

12  +  22  +  32  +  42  + +n\ 

/yi   I'll  _1_  1   I 

Here  6?  =  1,  an  =  n,  etc.,  /S^  =  1  +  2 +  /^  =  — -, 

so  that  (3)  gives 

_  {71  +  1)^  —  1  __  n{n-\-l)  _^  n 
^  ~  8  2  3* 

Noting  that  ;^  +  1  is  a  factor  of  the  second  member,  we 
may  reduce  this  equation  to 

^^^n{n  +  l)(2n±t)^  (4) 

which  is  the  required  expression  for  the  sum  of  the  squares  of 
the  first  n  numbers. 


S^  =  -^:7 dS^—-^  Sq,  (3) 


SERIES.  343 

393.  To  find  the  sum  of  the  cubes  of  any  progression, 
we  put  m  =  3  in  the  equation  (2),  which  then  gives 

^Li  —  ^f      3  1 

^^  ==        4^    -  i  ^'Sf,  -  d^S,  -  I  <P8,.  (6) 

Applying^  this  as  before  to  the  case  in  which  a^,  a^,  ct^, 
etc.,  are  the  natural  numbers,  1,  2,  3,  etc.,  we  find 

_  (^  +  1)4-,!       3  1 

^z  —  1 2    2  ~"  ^1  —  4  ^0 

_  {n  +  1)^  —  1  _  n{n  +  l)  {2n  +  1)  __  n{n  +  l)  _  n 
""  4  4  ""        2         ~"  4' 


Separating  the  factor  n  +  1  and  then  reducing,  this  equa- 
tion becomes 

2 

(5) 


But  — ^- — -  is  the  sum  of  the  natural  numbers 

1  +  2  4-  3  +  etc., 

and  S^  being  the  sum  of  the  cubes,  we  have  the  remarkable 
relation, 

13  4.  23  +  33  + 4.  ^^3  _  (1  _|.  2  +  3  + +  nf. 

That  is,  the  sum  of  the  cubes  of  the  first  n  numhers  is 
equal  to  the  square  of  their  sum. 

We  may  verify  this  relation  to  any  extent,  thus  : 
When  71^2,  13  +  23:=  1  +  8  =  9  =  (1  +  2)^ 
When  yi^3,  13+23  +  33  =  1  +  8  +  27  =  36  ==  (1  +  2  +  3)2. 
When  n  =  4.,  13  +  23  +  33  +  43  =  1+8  +  27  +  64  =  100  =  (1  +  2  +  3  +  4)2. 
etc.  etc.  etc.  etc. 

393.  Enumeration  of  a  Rectangular  Pile  of  Balls,  The 
preceding  theory  may  be  applied  to  the  enumeration  of  a  pile 
of  balls  of  which  the  base  is  rectangular  and  each  ball  rests  on 
four  balls  below  it.  Let  us  put  p,  q,  the  number  of  balls  in 
two  adjacent  sides  of  the  base. 


344  SERIES. 

Then  the  second  course  will  have  p  —1  and  q  —  1  balls 
on  its  sides ;  the  third  p  —  %  and  ^  —  2,  and  so  on  to  the  top, 
which  will  consist  of  a  single  row  of  j»  —  §'  +  1  balls  (suppos- 
ing p  >  q).  The  bottom  course  will  contain  jjq  balls,  the  next 
course  {p  —  1)  (^  —  1)>  etc.  The  total  number  of  balls  in  the 
pile  will  be 
i\rz:.;,^  +  (j,->l)(g-l)+(^-2)(^-2)  +  ....+(p-^  +  l).  (6) 

To  find  the  sum  of  this  series,  let  us  first  suppose  p^=^q, 
and  the  base  therefore  a  square.     We  shall  then  have 

N'  =  q^  -^  {q  -If  +  {q  -2f  +  .  .  .  .  -{-1, 
which  is  the  sum  of  the  squares  of  the  first  q  numbers. 
Therefore,  by  §  291,  (4), 

^,  ^  ^(^-fl)(2^+l)  ^^^ 

Next  let  us  put  r  for  the  number  by  which  p  exceeds  q  in 
the  general  expression  (6).     This  expression  will  then  become 

]V=  q{q  +  r)  +  {q-l){q-l-^r)  +  {q-2){q-^2-{-r)  + 

+  (1  +  r) 
==  ^'  -f  (^  -  1)'  +  (^  -2)2  +  .  .  .  .  +  22  +  1 

+  [^  +  (^-l)+(^-^)  +  ....  +1]^ 

=  l(lill)^.±i)  +  iil±l)  ,  (§  291,  4.) 

_q{q-{-l)  (3r  -{- 2q  +  1) 
~  6 

EXERCISES. 

1.  Find  the  sum  of  the  first  20  numbers,  1  +  2  +  3+  ...  . 
+  20,  then  the  sum  of  their  squares,  and  the  sum  of  their 
cubes,  by  successive  substitutions  in  the  general  equation  (2). 

2.  Express  the  sum  and  the  sum  of  the  squares  of  the  first 
r  odd  numbers,  namely, 

1  +  3   +  5  +....  +  (2/-  -  1), 
and  12  -f  32  +  52  + +  (2r  -  1)2. 

3.  Express  the  sum  of  the  first  r  even  numbers  and  the 
sum  of  their  squares,  namely, 

2  +  4  +  6  +  .  . .  ,  +  2r, 
and  22  +  42+62  + +(2r)2. 


8ERIE8,  345 

4.  A  rectangular  pile  of  balls  is  started  with  a  base  of  p 
balls  on  one  side  and  q  on  the  other.  How  many  balls  will 
there  be  in  the  pile  after  3  courses  have  been  laid  ?  How 
many  after  s  courses  ? 

5.  Find  the  value  of  the  expression 

S  (a  +  Z>a;  +  cx% 

6.  Find  the  value  of 

2  (a  +  te  +  cx^), 

x=l 

294.  To  find  the  sum  of  n  terms  of  the  series 
111  1 


1-2  ^  2.3  ^  3.4  ^  ^  n{n  H-  1) 

Each  term  of  this  series  may  be  divided  into  two  parts, 

thus : 

J___l_l  1    _  1       1 

1-2  ""  1       2'  2.3  ""2      3' 

1  11 


n{n  -{-  1)       n       n  -\-\ 
Therefore  the  sum  of  the  series  is 

in  which  the  second  part  of  every  term  except  the  last  is  can- 
celled by  the  first  part  of  the  term  next  following.  Therefore 
the  sum  of  the  n  terms  is 


If  we  suppose  the  number  of  terms  n  to  increase  without 

limit,  the  fraction will  reduce  to  zero,  and  we  shall  have 

^^  +  1 

jT^  +  273  +  3T4  +  ^^^">  ^^  i'y^finU'iim  —  1. 

This  is  the  same  as  tlie  sum  of  the  geometrical  progression,  s  +  7  +  o 

<o       4       o 


346  *  SERIES, 

4-  etc.,  ad  infinitum.    It  will  be  interesting  to  compare  the  first  few  terms 
of  the  two  series.     They  are 

3t      i      1      Jl      Jl      i. 
2  "^  6  "^  12  "^  20  "^  30  "^  42 ' 

1      i       1       Jl      i_      Jl 
2"^  4^  8  "^  16"*"  32"^  64' 

We  see  that  the  first  term  is  the  same  in  both,  while  the  next  three 
are  larger  in  the  geometrical  progression.  After  the  fourth  term,  the 
terms  of  the  progression  become  the  smaller,  and  continue  so. 

395,  Generalization  of  the  Preceding  Result,  Let  us  take 
the  series  of  which  the  n^^  term  is 

_P 


{i  ■\-  n  —  1)  (y  +  ^  —  1) 
The  series  to  n  terms  will  then  be 
I   .  I ,  I , 

a  ^  {i  + 1)  u  + 1)  ^  (^'  +  ^)  (y  +  2)  ^  •  •  •  • 


{i  -{-n  —  1)  (y  +  7^  —  1) 
If  we  suppose  j  >  i,  and  put,  for  brevity, 
h  =zj  —  i, 
the  terms  may  be  put  into  the  form 


ij       h  \i       jr 


ij         ic  \l        J/ 
P_ P  /  JL L\ 

-  jc  \i  +  1    _/  +  ir 


{^  + 1)  U  +  1)     ^  Vi*  +  1    y  + 

etc.  etc. 

P 


^li—1 1_). 

k\i  4-  n  —  1       1  -^n—V 


(i  -\-  n  —  1)  (y  +  /^  +  1)       k\i  -\-  n  —  1      j  -{-  n 

When  we  add  these  quantities,  the  second  part  of  each  term 
will  be  cancelled  by  the  first  part  of  the  ¥^  term  next  follow- 
ing, leaving  only  the  first  part  of  the  first  k  terms  and  the 
second  part  of  the  last  k  terms.     Hence  the  sum  will  be 

PJl  +  J_  +         +^_  _  J: ^!__.... L__). 

Ic\i       i-^1^    '         y+1       i-hn      i  +  n—1         j  +  n—1/ 


SERIES.  347 

Example.     To  find  the  sum  of  n  terms  of  the  series 
1  1  1.1.  .  1 


2. 

1         1/1 
3-6 


2-5   '   3.6    '   4.7   •   5-8 (n-\-l){n  +  4:) 

Each  term  may  be  expressed  in  the  form 

2-5        3\2       5/' 

!\3       6/' 

4.7       3  \4       7/' 

__!__=  1(1 1_) 

TC  (w  +  3)       3  Vw      w  +  3/' 

1 =  1  (_i ^v 

(?z  +  1)  (^  +  4)        3  \/^  +  1       ?^  +  4/ 

Therefore,  separating  the  positive  and  negative  terms,  we 
find  the  sum  of  the  series  to  be 


1/1111  1  1 

3\2       3       45  n      n  -\-l 


11  111 


_  __1 l\ 

5       6  n      n  +  1      n-{-2      n  +  3      ?^-f4/' 

or,  omitting  the  terms  which  cancel  each  other. 


1(1  +  1  +  1 1 i_ 

3\2^3^4       2^  +  2       u -\- 3       n -]- 


y- 


When  n  is  infinite,  the  sum  becomes 

3\2  ^3^4/        3  12       36 

EXERCISES. 

What  is  the  sum  of  n  terms  of  the  series : 

'-  3^ +  4^  +  5^  + ^*'^- 
j_   j_    j_  1 


3.5   '   5.7^7-9^ ^  {2n  +  l)(2n  +  3) 


348  8EBIE8- 

2 


3- 

4- 

1-3    '   2-4   '   3-5^ 

5- 

Sum  the  series 

1 

1 

■•■  (w  +  1)  (;^  +  4)' 


7^  (?i  +  2) 

^J^T)  +  («  +  1)  (a  +  2)  +  (a  +  2)  (a  4-  3)  +  ^^''  ""^  ^^^' 

396.    To  sum  the  series 

^  =  1  +  2r  +  3r2  +  4r3  +  etc. 

Let  us  first  find  the  sum  of  n  terms,  which  we  shall  call 
8n^     Then 

^^  ==  1  4-  2r  +  3r2  +  A.r^  + /^r~-l. 

Multiplying  by  r,  we  have 

r8n  =  r  -^^  2r2  +  Sr^  +  4r4  +  .  .  .  .  +  ^ir'*. 
By  subtraction, 

{l  —  r)8n  =  l-]-r-\-r^-\-7^ +  r^^-i  —  nr^ 

=  3 :^ir'*  (§  212,  Prob.  V). 

1  —  r  ^Q        y  / 

Therefore,       Sn  =  j. 


(1  ^ry      1  —  r 

Now  suppose  n  to  increase  without  limit.  If  r  >  1,  the 
sum  of  the  series  will  evidently  increase  without  limit. 

If  r  <  1,  both  7'^  and  nr^  will  converge  toward  zero  as  n 
increases  (as  we  shall  show  hereafter),  and  we  shall  have 

1 


S  = 


(1  -  ry 


EXERCISES. 

Find  in  the  above  way  the  sum  of  the  following  series  to  n 
terms  and  to  infinity,  supposing  r  <  1 : 

1.  G5  -f  Sar  -h  5ar^  -f  7ar^ ....  +  (2/i  —  1)  ar^-~^» 

2.  2fl5  4-  4ar  -f-  6ar^  +  Sar^ -f  2nar'^--\ 

3.  {a  +  J)  r  -f  (^  -f  2^)  r2  +  .  .  .  .  +  (a  +  /^&)  r^. 


SERIES.  349 


397.    Sum  the  series 

r^  +  ark  +  ra  + '*''•'  <«) 

of  which  the  general  term  is  _^_j-J^— ^-. 

Let  us  find  whether  we  can  express  this  series  as  the  sum 
of  two  series.    Assume 

n{n-\-l){n  -\-  2)  ^  n{n-\-l)  "^  (n  -}- 1)  {n  +  2) ' 

where,  if  possible,  the  values  of  the  indeterminate  coefficients 

A  and  B  are  to  be  so  chosen  that  this  equation  shall  be  true 

identically. 

Keducing  the  second  member  to  a  common  denominator, 

we  have 

1 {A  +  B)  n-\-2A 

n  {n  +  1)  {n  -f-  2)  ~  n  {n -\- 1)  {n -{-  2)' 

In  order  that  these  fractions  may  be  identically  equal,  we 
must  have 

{A  -]-  B)  n  +  2A  =z  1,  identically y 

which  requires  that  we  have  (§  281), 

A  -{-  B  =  0,        2A  =  1. 

This  gives  A  =  -,  i5  ==  —  -• 

Therefore, 

1  1111 


n{n  +  l){n-\-2)       2n{n-\-l)       2  {n  +  1)  {n  +  2)' 

60  that  each  term  of  the  series  (a)  may  be  divided  into  two 
terms.     The  whole  series  will  then  be 

We  see  on  sight,  that  by  cancelling  equal  terms,  the  sum  of 
n  terms  is  ^^        1  1 

On  =  -J 


4      2{n  +  l){7i  +  2)' 
and  the  sum  to  infinity  is  j* 


350  SERIES. 

398.  Consider  the  harmonic  series 

1  1  1  X 

1  +  ^  +  3+j  +  etc., 

of  which  the  oi^^  term  is  —  This  series  is  divergent,  because 
we  may  divide  it  into  an  unlimited  number  of  parts,  each 
equal  to  or  greater  than  -,  as  follows: 

1st  term  =  !,>-; 

2d  term  =  ^; 

3d  and  4th  terms  >  ^ ; 
etc.  etc. 

In  general,  if  we  consider  the  n  consecutive  terms, 

the  smallest  will  be  ^ ,  and  therefore  their  sum  will  be  greater 

than  :r-  X  ^,  that  is,  greater  than  -• 
Zn  Z 

]^ow  if  in  {a)  we  suppose  n  to  take  the  successive  values, 
i,  2,  4,  8,  16,  etc.,  we  shall  divide  the  series  into  an  unlimited 

number  of  parts  of  the  form  {a),  each  greater  than  ^-  There- 
fore, the  sum  has  no  limit  and  so  is  divergent. 

Of  Differences. 

299.  When  we  have  a  series  of  quantities  proceeding  ac- 
cording to  any  law,  we  may  take  the  difference  of  every  two 
consecutive  quantities,  and  thus  form  a  series  of  differences. 
The  terms  of  this  series  are  called  First  Differences. 

Taking  the  difference  of  every  two  consecutive  differences, 
we  shall  have  another  series,  the  terms  of  which  are  called 
Second  Differences. 

The  process  may  be  continued  so  long  as  there  are  any  dif- 
ferences to  write. 


SERIES.  351 

Example.  In  the  second  column  of  the  following  table 
are  given  the  seven  values  of  the  expression 

^  _  10a;3  ^  30^2  _  40:c  +  25  =  (px, 
for  X  =  0,1,  2,  3,  4,  5,  6. 

In  the  third  column  a'  are  given  the  differences, 
6  —  25  =  —  19,     1  —  6  :^  ~  5,     —  14  —  1  =  —  15,    etc. 

In  column  a"  are  given  the  differences  of  these  differences, 
namely, 

_  5  _  (_  19)  r=  +  14,         _  15  —  (-  5)  =  —  10,    etc. 

X  (j>X  A'  A''  A"'  Aiv  Av 

0  +25 

—  19 
1+6  +14 

—  5  —24 
2+1                   —10  +24 

—  15  0  0 
3—14                   —10                   +24 

—  25  +24  0 

4  _  39  +14  +24 

—  11  +  48 

5  _  50  +62 

+  51 

6  '+    1 

The  process  is  continued  to  the  fourth  order  of  differences, 
*  which  are  all  equal,  whence  those  of  the  fifth  and  following 
orders  are  all  zero. 

It  will  be  noted  that  the  sign  of  each  difference  is  taken  so 
that  it  shall  express  each  quantity  7ninus  the  quantity  next 
preceding.     We  have  therefore  the  following  definitions  : 

300.  Def.  The  First  Difference  of  a  function  of 
any  variable  is  the  increment  of  the  function  caused  by 
an  increment  of  unity  in  the  variable. 

The  Second  Difference  is  the  difference  between 
two  consecutive  first  differences. 
■    In  general,  the  n*^'  Difference  is  the  diflference  be- 
tween two  consecutive  {n  —  ly  diflferences. 


352  8EBIE8. 

To  investigate  the  relation  among  the  differences,  let  ua 
represent  the  successive  numbers  in  each  column  by  the  indices 
1,  'Z,  3,  etc.,  and  let  us  put  A^,  Ag,  A3,  etc.,  for  the  values  of 
^x.  We  shall  then  have  the  following  scheme  of  differences, 
in  which 

a;  =  Aj  —  Ao,     a;  =  Ag  -  Ai,     a;  =  A3  -  Ag  ; 

a;  =  a;-a;,   a';=:.a;-a;,   a;  =  a;-a;; 

a:  =  a;-a;,   a-  =  a;-a';,   a;:=a;-a- 

etc.  etc.  etc. 

the  w'*  order  of  differences  being  represented  by  the  symbol  A 
■with  n  accents. 

Ao 


Ai  A'; 

^3  •  a; 


A'  A"' 

^2  ^1  ^0 

/// 

1 


An-1 

Let  us  now  consider  the  following  problem : 

To  express  Ai  in  teriJis  0/  Aq,  Aq?  Aq?  etc. 
We  have,  by  the  mode  of  forming  the  differences, 
Ai  =z  Ao  +  a;,     a;  =  a'o  +  Ao,     Ai  =  A^  +  a'o,  etc.       {a) 

Ag  =  Ai  +  a;,     Ag  =:  Ai  +  Ai,     Ag  =  Aj  +  a'''   etc. 

If  in  this  last  system  of  equations,  we  substitute  the  values 
of  A^,  Ai,  etc.,  from  the  system  (a),  we  have 

Ag  =  Ao  +  2a;  4-  a;,  a;  =  a;  +  2a;  +  a';,  etc.  (h) 

Again, 

A3  ::=.  Ag  -h  a;,  a;  =  a;  +  a';,  a;  =  a;  +  Ag,  etc. 


DIFFERENCES.  353 

Substituting  the  values  of  Ag,  Ag,  etc.,  from  (5),  we  have 
A3  =  A„  +  3a'o  +    a; 

+  a;  +  2a;  +  a; 
or  A3  =  A„  +  3a;  +  3a;  +  a;'  (c) 


As 

=  A„  +  3a;  +  3a;  +  a;' 

A's 

=  a;  +  3a:+  a; 

+  a;  +  2a;'+a'j 

a;  =  a;  +  3a;  +  3a7  +  aj 

Forming  A^  =  A3  +  A3,  etc.,  we  see  that  the  coefficients 
of  Aq,  Ao,  etc.,  which  we  add,  are  the  same  as  the  coefficients 
of  the  successive  powers  of  x  in  raising  1  +  a;  to  the  rf^  power 
by  successive  multiplication,  as  in  §  171.  That  is,  to  form  A^, 
A'^,  etc.,  the  coefficients  to  be  added  are 

13  3    1 
13     3     1 

14  6    4    1 

and  these  are  to  be  added  in  the  same  way  to  form  A 5,  and  so 
on  indefinitely.  Hence  we  conclude  that  if  i  be  any  index,  the 
law  will  be  the  same  as  in  the  binomial  theorem,  namely, 

Ai  =  Ao  +  ^'a'o  +  (I)  a;  +  (I)  a;'  +  etc.  ) 

A;-A;+tA;  +  (|)A:  +  (|)A-  +  etc.) 

To  show  rigorously  that  this  result  is  true  for  all  values  of 
1,  we  have  to  prove  that  if  true  for  any  one  value,  it  must  be 
true  for  a  value  one  greater.  Now  we  have,  by  definition, 
whatever  be  ^, 

Ai+i  =  Ai  +  Ai,         Ai^i  =  A^  +  Ai',      etc. 
Hence,  substituting  the  above  value  of  A^  and  Ai, 

Ai+i  =  Ao  +  (i  + 1)  a;  +  [(I)  +  ^j  a; 

23 


354  8EB1E8. 

We  readily  prove  that 

(i)--=m. 
(i)+(i)=m. 

etc.  etc. 

Substituting  these  values  in  {e),  the  result  is  the  same  given 
by  the  equation  {d)  when  we  put  ^  +  1  for  L 

The  form  {c)  shows  the  formula  to  be  true  for  i  =  3. 

Therefore  ifc  is  true  for  i  =  4. 

Therefore  it  is  true  for  i  =z  5,  etc.,  indefinitely. 

EXAMPLES    AND     EXERCISES. 

I.  Having  given  Aq  =  7,  A^  =  5,  Aq  =  —  2,  and  A'",  A'", 
etc.  =:  0,  it  is  required  to  find  the  values  of  A^,  A^,  A3,  etc., 
indefinitely,  both  by  direct  computation  and  by  the  formula  (d). 

We  start  the  work  thus: 

The  numbers  in  column  A''  are  all 
equal  to  —  2,  because  A'''  =  0. 

Each  number  in  column  A'  after 
the  first  is  found  by  adding  A''  or  —  2 
to  the  one  next  above  it. 

Each  value  of  Aj  is  then  obtained 
from  the  one  next  above  it  by  adding 
the  appropriate  value  of  A^ . 

This  process  of  addition  can  be 
carried  to  any  extent.  Continuing  it 
to  i  =  10,  we  shall  find  Ajo  =  —38. 

Next,  the  general  formula  (d)  gives,  by  putting  A^  =  7, 
A'jj  =  5,  A'^  =  —  2,  and  all  following  values  =  0, 

A,  =  7 +  5^-3^1), 

and  the  student  is  now  to  show  that  by  putting  i  —  1,  i  =  2, 
etc.,  in  this  expression,  we  obtain  the  same  values  of  Ai,  Ag, 
A3,  ....  Aio,  that  we  get  by  addition  in  the  above  scheme. 

It  is  moreover  to  be  remarked  that  we  can  reduce  the  last 
equation  to  an  entire  function  of  i,  thus : 
Ai  =  7  +  6^*  -  i\ 


% 

Ai 

Ai 

At 

0 

7 

+  5 

1 

+  12 

+  3 

-% 

2 

+  15 

+  1 

—  2 

3 

etc. 

—  1 

—  2 

4 

etc. 

-2 

etc. 

etc. 

DIFFERENCES.  355 

2.  Having  given  A^  —  5,  A'^  =  —20,  A'^  =  —30, 
A'^'  =  +  9,  it  is  required  to  find  in  the  same  way  the  values 
of  Aj  to  A5,  and  to  express  Ai  as  an  entire  function  of  i  by 
formula  {d). 

3.  On  March  1,  1881,  at  Greenwich  noon,  the  sun's  longi- 
tude was  341°  5'  10". 9  ;  on  March  2  it  was  greater  by  1°  0'  9".0, 
but  this  daily  increase  was  diminishing  by  2"  each  day.  It  is 
required  to  compute  the  longitude  for  the  first  seven  days  of 
the  month,  and  to  find  an  expression  for  its  value  on  the  n^^^ 
day  of  March. 

4.  A  family  had  a  reservoir  containing,  on  the  morning  of 
May  5,  495  gallons  of  water,  to  which  the  city  added  regularly 
50  gallons  per  day.  The  family  used  35  gallons  oh  May  5, 
and  5  gallons  more  each  subsequent  day  than  it  did  on  the  day 
preceding.  Find  a  general  expression  for  the  quantity  of 
water  on  the  n*^  day  of  May ;  and  by  equating  this  expression 
to  zero,  find  at  what  time  the  water  will  all  be  gone.  Also  ex- 
plain the  two  answers  given  by  the  equation. 

Theorems  of  Differences. 

301.  To  investigate  the  general  properties  of  differences, 
we  use  a  notation  slightly  difibrent  from  that  just  employed. 

If  u  be  any  function  of  x,  which  we  may  call  ^Xy  so  that 
we  put 

U  =  (j)X, 

then  A^  =  0  (a;  -f  1)  —  <t)X,  {a) 

Here  the  symbol  A  does  not  represent  a  multiplier,  but 
merely  the  words  differ eiice  of. 

The  second  diJfference  of  ic  being  the  difference  of  the  dif- 
ference, may  be  represented  by  AA?^. 

For  brevity,  we  put 

A%  for  AA^, 

where  the  index  2  is  not  an  exponent,  but  a  symbol  indicating 
a  second  difference. 

Continuing  the  same  notation,  the  n^^  difference  will  be 
represented  by  A'^. 


356  SERIES, 

EXAMPLE. 

To  find  the  successive  differences  of  the  function 

u  =  ax^  +  Ix^, 

By  the  formula  {a),  we  have 

m  —  a(x  •\-  If  ^  h  {x  +  1)2  —  ax^  —  Ix^ ; 

and,  by  developing, 

l\u  —  Sax^  +  (3«  +  2b)x  -]-  a  -j-  b. 

Taking  the  difference  of  this  last  equation, 

^^u  =  3a  {x  +  1)2  +  {da  -{.  2b)  {x -i- 1)  +  a  +  b 

—  dax^  —  {3a-{-2b)x--a-'b 
=  6ax  -i-  6a  +  2b. 

Again  taking  the  difference,  we  have 

A%  =  6a{x  -\-  1)  —  6ax  =  6a. 

This  expression  not  containing  x,  A%,  A^u,  etc.,  all  vanish. 

EXERCISES. 

Compute  the  differences  of  the  functions : 

I.     x^  +  mx^  -ir  nx  -\-  p.  2.     2x!^  +  3x^  +  5. 

3.  hx^  +  lOx^  -t-  15. 

4.  In  the  case  of  the  last  expression,  prove  the  agreement 
of  results  by  computing  the  values  of  C^u^  /iht,  etc.,  for  x=:0, 
X  =  1,  and  a;  =  3,  and  comparing  them  with  those  obtained 
by  the  method  of  §  299.  The  latter  are  shown  in  the  follow- 
ing table : 

u  :=  5a^  +  lOx^  +  15. 

X         u  ^u  6?u  A^i* 

0 

15 

1  15 

30  50 

2  65  30 
95                           80 

3  145  30 
240                         110 

4  255 
495 

5 


DIFFERENCES.  857 

5.  Do  the  same  thing  for  exercise  2,  and  for  the  function 
tabulated  in  §  299. 

303.  It  will  be  seen  by  the  preceding  examples  and  exer- 
cises, that  for  each  difference  of  an  entire  function  of  x  which 
we  form,  the  degree  of  the  function  is  diminished  by  unity. 
This  result  is  generalized  in  the  following  theorem : 

The  n^^  differences  of  the  function  x^  are  constant 
and  equal  to  n ! 

Proof.  If  u  =  x^,  we  have,  by  the  definition  of  the  sym- 
bol ^, 

C^u  —  {x  -\-  lY  —  x% 

or  ^u  =z  7ix'^-'^  +  (1)^''"^  +  ^^^* 

That  is,  in  talcing  the  difference,  the  highest  power  of 
X  is  multiplied  by  its  exponent  and  the  latter  is  di7}%in- 
ished  hy  unity. 

Continuing  the  process,  we  shall  find  the  n^^  difference 
to  be     . 

n{n  —  l){n  —  2),,,.l  =  nl 

Cor,    If  we  have  an  entire  function  of  x  of  the  degree  n, 

ax^  +  Ix^-^  +  cx^-'^  -f-  etc., 

the  (n  —  \Y^  difference  of  hx^-^,  the  {n  —  2)^  difference  of 
c;r^"2,  etc.,  will  all  be  constant,  and  therefore  the  n^^  difference 
of  these  terms  will  all  vanish.  Therefore,  the  n^^  difference  of 
the  entire  function  will  be  the  same  as  the  n^^  difference  of 
ax^ ;  that  is,  we  have 

A^  [ax^  +  Ix^-"^  +  etc.)  =  an ! 

Hence,  the  n^^  difference  of  a  function  of  the  n^^  de- 
gree is  constant,  and  equal  to  n !  multiplied  hy  the  eoeffv- 
dent  of  the  highest  power  of  the  variable. 


358  SEEIE8, 

CHAPTER    IV. 

THE     DOCTRINE    OF     LIMITS. 

303.  The  doctrine  of  limits  embraces  a  set  of  principles 
applicable  to  cases  in  which  the  usual  methods  of  calculation 
fail,  in  consequence  of  some  of  the  quantities  to  be  used  van- 
ishing or  increasing  without  limit. 

We  have  already  made  extensive  use  of  some  of  the  princi- 
ples of  this  doctrine,  and  thus  familiarized  the  student  with 
their  application,  but  our  further  advance  requires  that  they 
should  be  rigorously  developed. 

Axiom  I.  Any  quantity,  however  small,  may  Ibe 
multiplied  so  often  as  to  exceed  any  other  fixed  quan- 
tity, however  great. 

Ax.  II.  Conversely^  any  quantity,  however  great, 
may  be  divided  into  so  many  parts  that  each  part  shall 
be  less  than  any  other  fixed  quantity,  however  small. 

Def,  An  Independent  Variable  is  a  quantity  to 
which  we  may  assign  any  value  we  please,  however 
small  or  great. 

Theoeem  I.  //  a  fraction  have  any  finite  numerator, 
and  an  independent  variable  for  its  denominator,  we 
may  assign  to  this  denominator  a  value  so  great  that 
the  fraction  shall  he  less  than  any  quantity,  however 
small,  which  we  may  assign. 

Proof,  Let  a  be  the  numerator  of  the  fraction,  x  its  de- 
nominator, and  a  any  quantity,  however  small,  which  we  may 
choose  to  assign. 

Let  71  be  the  number  of  times  we  must  multiply  a  to  make 
it  greater  than  a.     (Axiom  I.)     We  shall  then  have 

a  <  na. 

Consequently,  -  <  «. 


LIMITS.  359 

Hence,  by  taking  x  greater  than  n,  we  shall  have 

a  ^ 
-  <  cc. 

X 

Example.    Let  a  =  10.     Then  if  we  take  for  a  in  succes- 

'^"^'  m^  To;5oo'  imm^  ^^"-^  ^'  ^"^'  "^^y  ^^  ^"^^ 

aj  >  1,000,     X  >  100,000,    X  >  10,000,000,    etc., 
to  make  —  less  than  a, 

X 

In  the  language  of  limits,  the  above  theorem  is  expressed 
thus : 

The  limit  of  - ,  when  x  is  indefinitely  increased,  is 
zero. 

Theorem  II.  If  a  fraction  have  any  finite  numerator, 
and  an  independent  variable  for  its  denominator^  zve 
may  assign  to  this  denominator  a  value  so  small  that 
the  fraction  shall  exceed  any  quantity,  hoivever  great, 
which  we  may  assign. 

Proof.     Put  as  before  -  for  the  fraction,  and  let  A  be  any 

number  however  great,  which  we  choose  to  assign. 

Let  71  be  a  number  greater  than  A.  Divide  a  into  n  parts, 
and  let  a  be  one  of  these  parts  ;  then 

a  =  na. 


Consequently, 

a 

Therefore, 

if  we 

take  for  x  a  quantity  less  than  a,  we  shall 

have 

^>n>  A, 

X 

or 

i>'- 

Eem.     If  we  have  two  independent  variables,  x  and  y: 
We  may  make  x  any  number  of  times  greater  than  y. 


360  LIMITS, 

Then  we  may  make  y  any  number  of  times  greater  than 
this  value  of  x. 

Then  we  may  make  x  any  number  of  times  greater  than 
this  value  of  y» 

And  we  can  thus  continue,  making  each  variable  outstrip 
the  other  to  any  extent  in  a  race  toward  infinity,  without 
either  ever  reaching  the  goal. 

Theorem  III.  //  h  he  any  fixed  quantity,  however 
great,  and  a  a  quantity  ivhieh  zve  may  mahe  as  small 
as  lue  please,  lue  may  mahe  the  product  ha  less  than  any 
assignable  quantity. 

Proof,  If  there  is  any  smallest  value  of  ha,  let  it  be  s. 
Because  we  may  make  a  as  small  as  we  please,  let  us  put 

Multiplying  by  h,  we  find 

ha  <  s. 

So  that  ha  may  be  made  less  than  s,  and  s  cannot  be  the 
smallest  value. 

Def,  The  Limit  of  a  variable  quantity  is  a  value 
which  it  can  never  reach,  but  to  which  it  may  approach 
so  nearly  that  the  difference  shall  be  less  than  any 
assignable  quantity. 

Rem.  In  order  that  a  variable  X  may  have  a  limit,  it  must 
be  a  function  of  some  other  variable,  and  there  must  be  certain 
values  of  this  other  variable  for  which  the  value  of  X  cannot 
be  directly  computed. 

EXAMPLES. 

I.  The  value  of  the  expression 

^      x^  —  a^ 


X  —  a 


can  be  computed  directly  for  any  pair  of  numerical  values  of  x 
and  a,  except  those  values  which  are  equal.  If  we  suppose 
x^=ia,  the  expression  becomes 


LIMITS.  361 


0 


a  — a        0' 
which,  considered  by  itself,  has  no  meaning. 

2.  The  sum  of  any  finite  number  of  terms  of  a  geometrical 
progression  may  be  computed  by  adding  them.  But  if  the 
number  of  terms  is  infinite,  an  infinite  time  would  be  required 
for  the  direct  calculation,  which  is  therefore  impossible. 

3.  The  area  of  a  polygon  of  any  number  of  sides,  and  hav- 
ing a  given  apothegm,  may  be  computed.  But  if  the  number 
of  sides  becom.es  infinite,  and  the  polygon  is  thus  changed  into 
a  circle,  the  direct  computation  is  not  practicable. 

EXERCISE. 

^x 8 

If  we  have  the  fraction,  X  =  ~ ,  show  that  we  may 

7  1 

make  x  so  great  that  X  shall  differ  from  -  by  less  than  -— , 

less  than  _-^,  lesss  than  j^-^-,  and  so  on  indefinitely. 

Notation  of  the  Method  of  Limits. 

304.  Put  X,  the  quantity  of  which  the  value  is  to  be 
found ; 

X,  the  independent  variable  on  which  X  de- 
pends, so  that  Xis  a  function  of  ic; 

«,  the  particular  value  of  x  for  which  we  can- 
not compute  X; 

X,  the  limit  of  X,  or  the  value  to  which  it 
approaches  as  x  approaches  to  a. 

Then  the  limit  L  must  be  a  quantity  fulfilling  these  two 
conditions : 

1st.  Supposing  X  to  approach  as  near  as  we  please  to  a,  we 
must  always  be  able  to  find  a  value  of  x  so  near  to  a  that  the 
difference  L  -—  X  shall  become  less  than  any  assignable  quan- 
tity. 

2d.  X  must  not  become  absolutely  equal  to  L,  however 
near  x  may  be  to  a. 


362  '      LIMITS. 

Eem.     The  quantity  a,  toward  which  x  approaches,  may  be 
either  zero,  infinity,  or  some  finite  quantity. 

Example  i.     Suppose 

Y  —  ^  '^  ^ 
^  X  —  a 

By  §  93,  this*  expression  is  equal  to 

x^  +  ax  +  dj^,  {a) 

except  when  x:=za.     But  suppose  ^  to  be  the  difference  be- 
tween x  and  a^  so  that 

a;  =  «  +  (J. 

Substituting  this  value  in  the  expression  (a)^  the  equation 
becomes 

/TflS  __   /y3 

z=z  3a2  +  Sad  +  ^2. 

x  —  a 

Now  we  may  suppose  6  so  small  that  daS  -\-  6^  shall  be  less 
than  any  quantity  we  choose  to  assign.     Hence  we  may  choose 

ry'S  /vS 

a  yalue  of  x  so  near  to  a  that  the  value  of  shall  differ 

X  —  a 

from  3^2  by  less  than  any  assignable  quantity.     Hence,  if 

^  _  x^  —  a^ 
x  —  a  ^ 
then  L  =z  3a% 

oj3  __  ^3 

or  Sa^  is  the  limit  of  the  expression  as  x  approaches  a. 

X    — •    Ctr 
X 

Ex.  2.  The  limit  of  -,  when  x  becomes  indefinitely 

great,  is  unity. 

For,  subtracting  this  expression  from  unity,  we  find  the 
difference  to  be 

1 
X  +  1 

By  taking  x  sufficiently  great,  we  may  make  this  expression 
less  than  any  assignable  quantity.     (§  303,  Th.  I.)     Therefore, 

X 

approaches  to  unity  as  x  increases,  whence  unity  is  its 

X  -p  J. 

limit. 


LIMITS.  363 

Notation,  The  statement  that  L  is  the  limit  of  X  as  x 
approaches  a  is  expressed  in  the  form 

Lim.  X(a..=.a)  =  L. 

The  conclusions  of  the  last  two  examples  may  be  ex- 
pressed thus : 

^3  qZ  rt* 

Lim.  ix=a)  =z  Sa^,  Lim. (x=^)  =  1. . 

X  —  a  X  -\-  1 

Rem.  This  form  of  notation  is  often  used  for  the  follow- 
ing purpose.  Having  a  function  of  x  which  we  may  call  X, 
the  form  X(x=a)  means,  "  the  value  of  X  when  x:=  a." 

EXAMPLES. 

(x^  +  a)(x=:a)  =  a^  -\-  a.  {x^  —  a^)(x=^a)  =  0. 

If  we  require  the  limit  of  a  fraction  when  both  terms  be- 
come zero  or  infinite,  divide  both  terms  hy  some  commorb 
factor  which  hecomes  zero  or  infinity. 

Rem.  If  the  beginner  has  any  difficulty  in  understanding  the  pre- 
-  ceding  exposition,  it  will  be  sufficient  for  him  to  think  of  the  limit  as 
simply  the  value  of  the?  expression  when  the  quantity  on  which  it  de- 
pends becomes  zero  or  infinity. 

For  instance,  Lim.  ~    (a;  =  oo ), 

the  value  of  which  we  have  found  to  be  unity,  may  be  regarded  as  simply 
the  value  of  the  expression,  oo 

GO  +    1* 

Although  this  way  of  thinking  is  convenient,  and  generally  leads  to 
correct  results,  it  is  not  mathematically  rigorous,  because  neither  zero 
nor  infinity  are,  properly  speaking,  mathematical  quantities,  and  people 
are  often  led  into  paradoxes  by  treating  them  as  such. 

EXERCISES. 

Find  the  limit  of 

I. when  X  approaches  infinity. 

Divide  both  terms  by  x. 

2.  - — -—  when  X  approaches  infinity. 

uix? 

3.  — ^ when  X  approaches  infinity. 


364  LIMITS, 

\ X 

4.  z. —  when  X  approaches  infinity. 

/v2  r^ 

5.  ' when  X  approaches  a, 

X  —  €t 

6. when  x  approaches  infinity. 

a  —  x  ^^ 

Properties  of  Limits. 

305.  Theorem  I.  //  tivo  functions  are  equal,  they 
must  hajve  the  same  limit. 

Proyf.  If  possible^  let  Z  and  L  be  two  different  limits  for 
the  respective  functions.     Put 

SO  that  L  and  V  differ  by  2^;. 

Because  Z  is  the  limit  of  the  one  function,  the  latter  may 
approach  this  dmifc  so  nearly  as  to  differ  from  it  by  less  than  z. 

In  the  same  way,  the  other  function  may  differ  from  L 
by  less  than  z.  Then,  because  L  and  L  differ  by  %z,  the  func- 
tions would  differ,  which  is  contrary  to  the  hypothesis. 

Theorem  II.  The  limit  of  the  sum  of  several  func- 
tions is  equal  to  the  sum  of  their  separate  limits. 

Proof.     Let  the  functions  be  X,  X',  X",  etc. 
Let  their  limits  be     L,   L\  L'\  etc. 

Let  their  differences  from  their  limits  be  a,  a',  a'\  etc. 

Then  X  =  L  —  a, 

X'  =  L'  --  a\ 

X"  =  Z"  -  a", 

etc.  etc. 

Adding,  we  have 

.X+X'  +  X"  +  etc.  =  L-{-L'  +  r'-{-etc.-{a  +  a'  +  a''  +  etc.) 

The  theorem  asserts  that  we  may  take  the  functions  so  near 
their  limits  that  the  sums  of  the  differences  a -\- a' -\- a"  -^ etc, 
shall  be  less  than  any  quantity  we  can  assign. 


LIMITS.  365 

Let  h  be  this  quantity,  which  may  be  ever  so  small ; 
n,  the  number  of  the  quantities  a,  a',  a",  etc. ; 
a,  the  largest  of  them. 

Because  we  can  bring  the  functions  as  near  their  limits  as 
we  please,  we  may  bring  them  so  near  as  to  make 

k 
a  <  -  y         or         7ia  <  h 
n 

Then    «+«'-j-a"4-etc.  <  na  (because  a  is  the  largest) ; 

whence,  «  +  cc' +  «"  4- etc.  <  h. 

Therefore  the  sum  X+X'  +  JT^  +  etc.  will  approach  to 
the  sum  L  -\-  L'  +  L"  -^  etc.,  so  as  to  differ  from  it  by  less 
than  L  Because  this  quantity  Jc  may  be  as  small  as  we  please, 
L-{-L'-^L"-{-  etc.  is  the  limit  of  X+  X'  +  X"  +  etc. 

Theorem  III.  The  limit  of  the  product  of  two  func- 
tions is  equal  to  the  product  of  their  limits. 

Proof,  Adopting  the  same  notation  as  in  Th.  II,  we  shall 
have 

XX'  =  LL'  -aU  -a'L  +  aa'. 

Because  L  and  L'  are  finite  quantities,  we  may  take  a  and 
«'  so  small  that  aL'  -^a' L — aa  shall  be  less  than  any  quan- 
tity we  can  assign.  Hence  XX'  may  approach  as  near  as  we 
please  to  LL',  whence  the  latter  is  its  limit. 

Cor.  1.  The  limit  of  the  product  of  any  number  of 
functions  is  equal  to  the  product  of  their  limits. 

Cor.  2.  The  limit  of  any  power  of  a  function  is  equal 
to  the  power  of  its  limit. 

Theorem  IY.  TJ%e  limit  of  the  quotient  of  tiuo  func- 
tions is  equal  to  the  quotient  of  their  limits. 

Proof,  Using  the  same  notation  as  before,  we  have  for  the 
quotient  of  the  functions, 

X'  L'  —  a 

u 

while  the  quotient  of  their  limits  is  -^-^ 


366  LIMITS. 

The  difference  between  the  two  quotients  is 

L'        U  —  ex!         La  —  La 
L  L  —  a  L{L  —  a) 

If  L  is  different  from  zero,  we  may  make  the  quantities  a 
and  a'  so  small  that  this  expression  shall  be  less  than  any 

quantity  we  choose  to  assign.     Therefore,     -  is  the  limit  of 

L  —  a  .        ^  A. 

—p ,  that  IS,  OT  ^^' 

L  —  a  X 

jQn qti 

306.  Problem.     To  find  the  limit  of as  x 

7  X  —  a 

approacries  a. 

Case  I.     When  n  is  a  positive  whole  number. 

We  have  from  §  93,  when  x  is  different  from  a,  A 

x^ a"'  ^ 

x  —  a 

Now  suppose  X  to  approach  the  limit  a.  Then  x'^-'^  will 
approach  the  limit  a^"'^,  x^~^  the  limit  a^~%  etc.  Multiplying 
hy  a,a^,  etc.,  we  see  that  each  term  of  the  second  member 
approaches  the  limit  a^~K  Because  there  are  n  such  terms, 
we  have 

^.      x^  —  a^ 

Lim.  — — : —  (x=a)  =  na^~\ 
X  —  a 

Case  II.     WJien  n  is  a  positive  fraction. 

Suppose  n  =i~,  p  and  q  being  whole  numbers.    Then 

x^  —  a'^       x^  —  a^ 


x  —  a  X  —  a 

Let  us  put,  for  convenience  in  writing, 

x"^  =z  y,  (f  =z  h'^ 

then  X  ■=  'f,  a  —  1)^\ 

x^  —  a^  _y^  —  If  _     y  —  b 
1^^^^  ~  '"f^^^^  ~  "FEZ ' 
y-b 


LIMITS.  367 

As  X  approaches  indefinitely  near  to  a,  and  consequently  y 
to  b,  the  numerator  of  this  fraction  (Case  I)  approaches  to 
pl)p~^  as  its  limit  and  the  denominator  to  ql)Q~^.  Hence,  the 
fraction  itself  approaches  to 

phP-^  _  p 


qb^-^       q 


bP-^, 


Substituting  for  b  its  value  a^,  we  have 

Lim.  — ^^  ix^a)  =  ^  bP-Q  =  ^-a?  =  ^  J~' 
X  —  a  q  q  q 

Hence  the  same  formulae  holds  when  n  is  a  positive  fraction. 

Case  III.     When  n  is  negative. 

Suppose  n  =  —p,p  itself  (without  the  minus  sign)  being 


supposed  positive.    Then 

\  X  ^  a  / 


= =  x-P  arP 


—  —  x~P  a'P  - 


X  —  a 


When  X  approaches  a,  then    x~p    approaches    arP,    and 

a^  —  a^ 

approaches  ^a^~^     Substituting  these  limiting  values, 

X  d 

we  have 

X^  nj^ 

Lim. {x=a)  z=z  —  a'^PpaT^^  =  —  parP^K 

X  —  a  ^  ^ 

Substituting  for  —  p  its  value  /^,  we  have 

/jfjfl   /77l 

Lim. (cc=a)  =  na^~^. 

X  ^  a 

Hence, 


Theorem.    The  formulce 


x^  —  oP-  ^  . 

Lim. ix=d)  =  naP''^ 

X  --  a 


is  true  for  all  values  of  n,  whether  entire  or  fractional, 
positive  or  negative. 


368  BINOMIAL    TUEOUEM 

CHAPTER    V. 
THE    BINOMIAL    AND    EXPONENTIAL    THEOREMS. 


i 


The  Binomial  Theorem  for  all  Values  of  the 
Exponent. 

307.  We  have  shown  in  §§  171,  264,  how  to  develop 
{l-\-xY  when  ^  is  a  positive  whole  number.  We  have  now  to 
find  the  development  when  n  is  negative  or  fractional.    Assume 

(1  +  xY  =  -^0  +  B^x  +  B^x^  +  B^a^  +  etc.,         {a) 

Bq,  B^y  etc.,  being  indeterminate  coefficients.  Because  this 
equation  is  by  hypothesis  true  for  all  values  of  x,  it  will  remain 
true  when  we  put  another  quantity  a  in  place  of  x.     Hence, 

(1  -{-aY  =  B^  -i-  B^a  +  B^a?  +  B^a^  +  etc.  {b) 

Subtracting  {V)  from  (a),  and  putting  for  convenience 

X     =Z       1      -{-     X,  A       =       1      -\-     CCy 

the  difference  of  'the  two  equations  (a)  and  (b)  will  be 
X''-A''  =  B^  {x  -  a)  +  B.,  {x^  -  a^)  +  B^  {x^  -  a^)  +  etc. 

The  values  we  have  assumed  for  Xand  A  give 

X—  A  =:  X  —  a. 

Dividing  the  left-hand  member  by  X --  A,  and  the  right- 
hand  member  by  the  equal  quantity  x  —  a,  we  have 

-X^TX-  =  ^1  +  ^.  -^ZTa  +  ^3  ^— ^  +  etc. 

Now  suppose  X  to  approach  a.     The  limit  of  the  left-hand 
member  will  be  nA^'K     Taking  the  sum  of  the  corresponding* 
limits  of  fche  right-hand  member,  we  shall  have 

nA%-^  =  B^  +  2B^a  +  dB^a^  -f  4.B^a^  +  etc. 

Eeplace  A  by  its  value,  1  +  a,  and  multiply  by  1  +  a^ 
We  then  have 


BINOMIAL    THEOREM.  369 

^(1  +  aY  =  ^1  (1  +  «)  +  2^3«(1  +  «)  +  3Z?3a2(l  +  a) 

+  4i?4a3  (1  _|.  ^)  +  etc. 

=  J5i  +  (^,  +  2i?2)  a  +  i2B,  +  3^3)  a^ 

+  (3^3  +4i?Ja3  4-eto. 

Multiplying  the  equation  (b)  by  ^,  we  have 

/2  (1  +  a)^  =  Ti^o  +  ^-^1^  +  ^^2^2  +  iiB^a^ 

Equating  the  coeflBcients  of  the  like  powers  of  a  in  these 
equations  (§  281),  we  have,  first, 

B^  =  7iB^. 
By  putting  a  =  0  in  equation  (&),  we  find  Bq  =  1,  whence 

Then  we  find  successively, 
2B^  =  (n-l)  Bj,  whence  B^  =  -^  B^  =  !i(^ll. 

3B,  =  in-2)B,,       "        S3='^'^.=^^^|='-^. 

Substituting  these  values  of  Bq,  B^,  B^,  etc.,  in  the  equa- 
tion {a)  and  using  the  abbreviated  notation,  w^e  obtain  tho 
equation 

(1  +  xY  =  l  +  nx-\-(^'^)x^  +  (I)  x^  +  etc.,         (c) 
which  equation  is  true  for  all  values  of  n, 

308.  There  is  an  important  relation  between  the  form  of 
this  development  when  niQ  a,  positive  integer,  as  in  §§  171  and 
264,  and  when  it  is  negative  or  fractional.  In  the  former 
case,  when  we  form  the  successive  factors  n  —  1,  n  —  2, 
u  —  3,  etc.,  the  ?i^^  factor  will  vanish,  and  therefore  all  the 
coefficients  after  that  of  x^  will  vanish. 

But  if  n  is  negative  or  fractional,  none  of  the  factors 
n  —  1,  71  —  2,  etc.,  can  become  zero,  and,  in  consequence,  the 
series  will  go  on  to  infinity.  It  therefore  becomes  necessary, 
in  this  case,  to  investigate  the  convergence  of  the  development. 

If  i«;  >  1,  the  successive  powers  of  x  will  go  on  increasing 

indefinitely,  while  the  coefficients  (-),   (^),  etc.,  will  not  go 
24 


370  BINOMIAL    THEOEEM. 

on  diminisliing  indefinitely  in  the  same  ratio.  For,  let  us 
consider  two  successive  terms  of  the  development^  the  (^  +  1)*^, 
and  the  {i  4-  2)"^,  namely. 


The  quotient  of  the  second  by  the  first  is 

(n     \         /n\       n  —  i 


As  I  increases  indefinitely,  tliis  coefficient  of  x  will  approach 
the  limit  —  1  (§  304),  while  x  is  by  hypothesis  as  great  as  1. 
Therefore,  by  continuing  the  series,  a  point  will  be  reached 
from  which  the  terms  will  no  longer  diminish.     Therefore,       | 

The  development  of  {I  +  xy^  in  powers  of  x  is  not  con- 
vergent unless  a;  <  1.  j 

In  consequence,  if  we  develop  {a  -f  b)^  when  n  is  negative 
or  fractional,  we  must  do  so  in  ascending  powers  of  the  lesser 
of  the  two  quantities,  a  or  b, 

EXAM  PLES. 

I.  Develop  (1  +  x)^,  or  the  square  root  of  1  +  a;. 
Putting  7Z  =  ^,  we  have 


©  =  r 


(1)  = 


2V2-')  M 


1/1 

2_ 
1.2        ^  ""2.4' 

1.2.3  ~"  2.4.6' 

-      ^-  1.1.3.5 


in\  __  2 /^\  _  _  ll 

W  ""      4     \3/  ""       2T4.6.8 
etc.  etc.  etc. 


BINOMIAL    THEOREM.  371 

Whence, 
/■.    ,     ^i       1,1  1-1^  .  1-1-3    ,      1-1-3.5    ,  ,     ^ 

If  iz;  is  a  small  fraction,  the  terms  in  x^,  a^,  etc.,  will  be 

much  smaller  than  -  x  itself,  and  the  first  two  terms  of  the 

series  will  give  a  result  very  near  the  truth.     We  therefore 
conclude : 

Tl%e  square  root  of  1  plus  a  small  fraction  is  approxi- 
mately equal  to  1  plus  half  that  fraction. 

2.  To  develop  a/10. 

We  see  at  once  that  VlO  is  between  3  and  4.     We  put  10 
in  the  form 

3^  +  1=^32(1  +  1), 

when  VlO  =  3  (l  +  ^^ 

Then,  by  the  development  just  performed, 

l28""9^  +  ^^''- 


(l+J)    -l  +  2'9- 

.     1-4-      1 
8.92       16-93 

We  now  sum  the  terms  : 

1st  term. 

1.0000000 

2d      " 

=  1st  ~  18,  ...    . 

+   .0555556 

3d      " 

:::3    2d     ^    -  36,    .      .      . 

-  .0015432 

4th     " 

:=  3d    -^ 18,  .    .    . 

+  .0000857 

5th     " 

=  4th   X  —  5-r-72,     . 

—  .0000060 

6th     " 

=  5th  X  —7-^90,     . 

+  .0000005 

Sum  =  (1  +  ^ 

=  1.0540926 

lence, 

VlO  r=  3  X  sum 

=  3.1622778 

which  may  be  in  error  by  a  few  units  in  the  last  place,  owing 
to  the  omission  of  the  decimals  past  the  seventh. 


372  EXPONENTIAL    THEOREM, 

3.  To  develop  VS. 

We  see  that  3  is  the  nearest  whole  number  of  the  root.    So 
we  put 

V8  =  V(3^-l)=^3^(l-J-)=3(l-J)*, 
from  which  the  development  may  be  effected  as  before. 

EXERCISES. 

1.  Compute  the  square  root  of  8  to  6  decimals,  and  from,  it 
find  the  square  root  of  2  by  §  183. 

2.  Develop  (1  —  x)^. 

3.  Develop  (1  —  x)~^y  and  express  the  term  in  xK 

^       1  1-3   ,       1.3.5   , 

•    ^+2^  +  2:4^+2:4:6'^^  +^^^- 

Term  in  cc*  = 


2.4.6 M 

4.  Develop y  and  express  the  general  term. 

(1  +  xy 

5.  Develop  (1  +  -)    and  express  the  general  term. 

6.  Develop  (1  —  xf,  and  express  the  general  term. 

7.  Develop  the  m^^  root  oi  1  -{-  m, 

8.  Develop  (a  —  b)~%  when  a  <C  b, 

9.  Develop  (1  —  x)-^,  when  x  y  1, 

Because  the  development  will  not  be  convergent  in  ascend- 
ing powers  of  x  when  x  >  1,  we  transform  thus : 

!-.=  _. (l-l), 

/         IV 
and  so  put        (1  —  x)-^  =  (—  x)~^  (1 1 


xt 


10.  Develop  the  wP^  power  of  1  H 

11.  Compute  the  cube  root  of  1610  to  six  decimals. 


EXPONENTIAL    THEOREM,  373 

12.  Develop  {Va  +  Viy, 

13.  Using  the  functional  notation, 

0  (m)  =  1  +  (^):,  +  (1)^  +  g),,3  +  etc, 

multiply  the  two  series,  0  (m)  and  0  (^),  and  show  by  the  for- 
mulas of  §  261  that  the  product  is  equal  to  0  (m  +  n). 

The  Exponential  Theorem. 

309.  Let  it  be  required,  if  possible,  to  develop  a^  in 
powers  of  a;,  a  being  any  quantity  whatever.     Assume 

a""  =  Cq  +  C^x  +  C^x^  +  C^x^  +  etc.  (1) 

to  be  true  for  all  values  of  x.     Putting  any  other  quantity  y  in 
place  of  x^  we  shall  have 

ay  =  G,  +  C^y  +  C,y^  +  G^y^  +  etc.  (2) 

By  the  law  of  exponents  we  must  always  have 

a""  X  av  =:  tt-^+y. 

Now  the  value  of  a^'^v  is  found  by  writing  x  -{-  y  for  x  in 
(1),  which  gives 

a^^y=  C,  +  G^{x+y)  +  C^{x  +  yf+C^{x^yY^QtQ.     (3) 

On  the  other  hand,  by  multiplying  equations  (1)  and  (2), 
we  find 

a^av  =  G,^  +  G^G^y  +  G^G^y"^  +  G.G^f    +  etc. 
+  (7o  6>  +     G^^xy  +  6\  (72^^  _|_  etc 

+  (7o^2^'  +  C^i^2^^^  +  etc.  ^  ^ 

+  ^0^3^^    +  etc. 

By  §  285,  the  coefficients  of  all  the  products  of  like  powers 
of  ir  and  y  must  be  equal.  By  equating  them,  we  shall  have 
more  equations  than  there  are  quantities  to  be  determined, 
and,  unless  these  equations  are  all  consistent,  the  development 
is  impossible.  To  facilitate  the  process  of  comparison,  we 
have  in  equation  (4)  arranged  all  terms  which  are  homogeneous 
in  x  and  y  under  each  other. 


374  BINOMIAL    THEOBEM. 

By  putting  x  —  0  in  (1),  we  find 

aO  ::::,    Q^^  wlienCG  C^    =  1.       (§  103.) 

Comparing  the  terms  of  the  first  degree  in  x  and  y  in  (3) 
and  (4),  we  find 

Coefficient  of  x,  C^  —  Cq  C-^  ; 

"  y,  C,  -  C,G,. 

These  two  equations  are  the  same,  and  agree  with  C^  =  1 ; 
but  neither  of  them  gives  a  value  for  C\,  which  must  therefore 
remain  undetermined. 

Comparing  the  terms  of  the  second  degree,  we  find,  by  de- 
veloping [x  +  yY, 

C^  {x^  +  2xy  +  2/^)  ==  C,x^  +  G^'xy  +  C^y\ 
which  gives  ^^2  =  ^V^ 

whence  (7o  =z  -—  (7.^. 

^        1-2    ^ 

Comparing  the  terms  of  the  third  order  in  the  same  way, 
we  have 

C^{x^^^x^y  +  Zxy^  +  y^)  =  C,a^-\-C,C^x^y  +  C^C^xy^+C,y^ 

which  gives  ^0^  =:  0^0^  z=  -C^^; 

<i 

whence  G^  =  j-^-  C^^ 

If  the  successive  values  of  C  follow  the  same  law,  we  shall 
have 

and  in  general,  C^  =  —^  C.^.  (5) 

n\ 

Let  us  now  investigate  whether  these  values  of  C  render 
the  equations  (3)  and  (4)  identically  equal. 

Let  us  consider  the  corresponding  terms  of  the  n^^  degree, 
n  being  any  positive  integer.     In  (3)  this  term  will  be 

Cn  {x  -f-  yy. 


EXPONENTIAL    THEOREM.  375 

Expanding,  it  will  be 


(^n 


x^  +  nx^-'^y  +  f  I)  x^-'^f'  +  i^\  x'^-^'f  +  etc.        (6) 
In  (4)  the  sum  of  the  corresponding  terms  will  be,  putting 

CnX^^G^ Cn-1  ^-^y-^  G^ Gn-%x^-^y^-\-  G^ Gn-^x^-^y^-^-^ia,  (7) 

The  first  terms  in  the  two  expressions  are  identical. 
The  comparison  of  the  second  terms  gives 

G 
nGn  —  G.Gn-v,        whence        Gn  —  -^  Gn-i. 

n 

This  corresponds  with  (5),  because  (5)  gives 

and  if  we  substitute  this  value  of  Gn-\  in  the  preceding  ex- 
pression for  Gn,  it  will  become 

n     _  ^___^ 

^''--  n{n-l)\~   7i\ 
which  agrees  with  (5). 

The  third  terms  of  (6)  and  (7)  being  equated  give 

G2  Gn-2- 


(1)^'. 


Substituting  the  values  of  Gn,  G^,  and  Gn-^  assumed  in  the 
general  form  (5),  we  have 

and  we  wish  to  know  if  this  equation  is  true. 

Multiplying  both  sides  by  n\  and  dropping  the  common 

factor  Gu  it  becomes 

(n\  n\ 

2)  -  2!  (^-2)1' 
which  is  an  identical  equation. 

In  the  same  way,  the  comparison  of  the  following  terms  in 
(6)  and  (7)  give 

ln\  __  n\  hi\  ___  7z! 


376  EXPONENTIAL    THEOREM. 

all  of  which  are  identical  equations.  Hence  the  conditions  of 
the  development,  namely,  that  (6)  and  (7),  and  therefore  (3) 
and  (4),  shall  be  identically  equal,  are  all  satisfied  by  the  values 
of  the  coefficients  C  in  (5).  Substituting  those  values  in  (1), 
the  development  becomes 

ax^l^  C^x  +  ~  C,^x^  +  ^-i-^  C^hc^  -f  etc.         (8) 

This  development  is  called  the  Exponential  Theorem, 
as  the  development  of  {a  +  hY  is  called  the  binomial  theorem. 

310.  The  value  of  (7^   is  still  to  be  determined.     To  do 

this,  assign  to  x  the  particular  value  -^  •  Then  the  equation 
(8)  becomes    .  ^ 

«^.  ^  1  +  1  +  -1^  +  _1_  +  _l^-  +  etc.,  ad  inf.     (9) 

The  second  member  of  this  equation  is  a  pure  number, 
without  any  algebraic  symbol.  We  can  readily  compute  its 
approximate  value,  since  dividing  the  third  term  by  3  gives 
the  fourth  term,  dividing  this  by  4  gives  the  fifth,  etc.     Then 

1  +  1  =  2.000000 

l-^1.2:rr  .500000 

1^1.2.3=  Ammi 

1  _^  1.2.3.4  =  .041667 

1  ^  1.2.3.4.5  =:  .008333 

1  -T-  1.2.3.4.5.6  =  .001389 

1  ^  1.2.3.4.5.6.7  =  .000198 

1  -^  1.2.3.4.5.6.7.8  =  .000025 

1  ^  1.2.3.4.5.6.7.8.9  =  _J0OOOO3 

Sum  of  the  series  to  6  decimals,  2.718282 

This  constant  number  is  extensively  used  in  the  higher 
mathematics  and  is  called  the  JSfaperian  hase.^  It  is  repre- 
sented for  shortness  by  the  symbol  e,  so  that  e  =  2.718282.... 

The  last  equation  is  therefore  written  in  the  form 

a^^  —  e, 
*  After  Baron  Napier,  the  inventor  of  logarithms. 


EXPONENTIAL    THEOREM.  377 

Eaising  to  the  C^^^  power,  we  have  a  =  e^\     Hence  : 

The  quantity  G^  is  the  exponent  of  the  power  to  which 
we  must  raise  the  constant  e  to  produce  the  jiumher  a. 

We  may  assign  one  vahie  to  a,  namely,  e  itself,  which  will 
lead  to  an  interesting  result.  Putting  «  =:  6,  we  have  G^  =1, 
and  the  exponential  series  gives 

+  etc.        (10) 

If  we  put  x=:l,  we  have  the  series  for  e  itself,  and  if  w(? 
put  X  z=:  —1,  we  have 

^=^  =  ^-^  +  172-1:273  +  iT2:3:4-^^^- 

We  thus  have  the  curious  result  that  this  series  and  (9)  are 
the  reciprocals  of  each  other. 

EXE  RCI  SES. 

1.  Substitute  in  the  first  four  or  five  terms  of  the  expres« 
sions  (6)  and  (7)  the  values  of  G2,  G^,  Gn-2,  etc.,  given  by  (5), 
and  show  that  (6)  and  (7)  are  thus  rendered  identically  equal. 

Note.  This  is  merely  a  sliprht  modification  of  the  process  we  have 
actually  followed  in  comparing  the  coefficients  of  like  powers  of  x  and  y 
in  (6)  and  (7). 

2.  Compute  arithmetically  the  values  of  2.71832,  2.7183-^ 
and  2.7183~2^  and  show  that  they  are  the  same  numbers,  to 
three  places  of  decimals,  that  we  obtain  by  putting  x  =  2, 
x=:  —1,  and  X  z=  —  2  in  (10),  and  computing  the  first  eight 
or  ten  terms  of  the  series. 

3.  Since  e^+*  =  ee^,  the  equation  (10)  gives,  by  substituting 
the  developments  of  e^^^  and  e^, 

,       ,  (1  +  xY       (1  +  xY       (1  -{-xV        ^ 


='( 


x^       x^       x^ 
^  +  "^  +  2!  +  3!  +  4!  +  ''^- 


It  is  required  to  prove  the  identity  of  these  developments 
by  showing  that  the  coefficients  of  like  powers  of  a;  are  equal 


378  L0OABITEM8. 

CHAPTER    VI. 

LOGARITHMS. 

311.  To  form  the  logarithm  of  a  number,  a  constant  num- 
ber is  assumed  at  pleasure  and  called  the  base, 

Def.  The  Logarithm  of  a  number  is  the  exponent 
of  the  power  to  which  the  base  must  be  raised  to  pro- 
duce the  number. 

The  logarithm  otx  is  written  log  x. 

Let  us  put  a,  the  base  ; 

X,  the  number ; 

I,  the  logarithm  of  x. 
Then  d  z=z  x, 

Eem.  For  every  positive  value  we  assign  to  x  there  will  be 
one  and  only  one  value  of  /,  so  long  as  the  base  a  remains  un- 
changed. 

Def.  A  System  of  Logarithms  means  the  loga- 
rithms of  all  positive  numbers  to  a  given  base.  The 
base  is  then  called  the  base  of  the  system. 

Properties  of  Logarithms. 

313.  Consider  the  equations, 

aO=  1;   I  Hogl   =0; 

a^  z=z  a\   \  whence  by  definition,  \  log  «j  =  1 ; 

a?  =  d?',)  (  log  a^  =  2. 

Hence, 

I.   The  logarithm  of  1  is  zero,  whatever  be  the  base, 
II.   The  logarithm  of  the  base  is  1. 

III.  The  logarithm  of  any  number  between  1  and  the 
base  is  a  positive  fraction. 

IV.  The  logarithms  of  powers  of  the  base  are  integers, 
but  no  other  loga^rithms  are. 


L0GABITHM8,  379 


Again  we  have 


«~^  =  - ;    \  /  log  -    =  —  1 ; 

a~^  =  -^ ;  >  whence  by  definition,  (  lo2r  -„  =  —  2 ; 
1     I  111 

Hence, 

V.  The  logarithm  of  a  nujnher  between  0  mid  1  is 
negative. 

Again,  as  we  increase  ?^,  the  value  of  a'^  increases  without 

limit,  and  that  of  —  approaches  zero  as  its  limit.     Hence, 

VI.  The  logarithm  of  0  is  negative  infinity. 

VII.  Theorem.  T1^e  logarithm  of  a  product  is  equal 
to  the  sum  of  the  logarithm's  of  its  factors. 

Proof.     Let  p  and  q  be  two  factors,  and  suppose 

h  =  log  p,  k  =  log  q. 

Then  a^  =^  P^  a^  =:  q. 

Multiplying,  a^a^  =  a^^^  =  pq. 

Whence,  by  definition, 

h  +  k  =  log  (pq), 
or  log  p  +  \ogq  z=  log  (pq). 

The  proof  may  be  extended  to  any  number  of  factors. 

VIII. '  Theorem.  The  logarithm  of  a  quotient  is  found 
by  subtracting  the  logarithm  of  the  divisor  from  that  of 
the  dividend. 

Proof.  Dividing  instead  of  multiplying  the  equations  in 
the  last  theorem,  we  have 

-,  —  a^-^  =  -• 
a*  q 


380  LOGARITHMS. 

Hence,  by  definition,        h  —  k=z\og^y 
or  logj;— log^  =  log^. 

IX.  Theorem.  The  logarithm  of  any  power  of  a  num- 
her  is  equal  to  the  logarithm  of  the  numher  multiplied 
by  the  exponent  of  the  power. 

Proof.    Let  h  =  log  ij,  and  let  n  be  the  exponent. 
Then  ofi'  =:  p. 

Raising  both  sides  to  the  n^^  power. 

Whence  nh  =  log^'^, 

or  n  log  p  •=.  log  p^, 

X.  Theorem.  The  logarithm  of  a  root  of  a  numher 
is  equal  to  the  logarithm  of  the  numher  divided  hy  the 
index  of  the  root. 

Proof.  Let  s  be  the  number,  and  let  p  be  its  n^^  root,  so 
that 

p  =  \^s        and        s  =  p^. 
Hence,  log  s  =  log  p^  =  7i  log  p.    (IX.) 

logs 


Therefore,  log^  = 


n 


,       n/         lo2r  s 
or  log  Vs  =  — -- — 


Note.    We  may  also  apply  Th.  IX,  since  p  =  A     Con- 


sidering -  as  a  power,  the  theorem  gives 
^ogp  =  -logs. 


EXERCISES. 

Express  the  following  logarithms  in  terms  ol  hgp,  log  g^ 
log  X,  and  logy,  a  being  the  base  of  the  system :  1 


LOGARITHMS.  381 

1.  Log  p%  Arts.  2  log^  +  log  q, 

2.  Log  2^^^' 

3.  Log^Y-  4-  l^og  pqVy^. 

5.  Log  -  =  log  xp~\  and  explain  the  identity. 

6.  Log||  =  logxyp-^q-\ 

Ans.    Log  X  +  log  y  —  logp  —  log  ^. 


7-  Log^,. 

8-  Log  ^^3- 

9.  Log  \/ic. 

10.  Log  Vx  Vy* 

II.  Logy|. 

12.  Log  V«. 

13.  Logaa;. 

14.  Log-. 

T         ^ 

IS.  Log-. 

|yj.     I^^^g  («^  +  ^)  +  log  («  —  aj) 

17.  Log  '\/a^  —  x\ 

18.  LogVl-«^. 

19.  Log  (a^  —  ^-2), 

Logarithmic  Series. 

313.  Rem.  The  logarithm  of  a  number  cannot  be  deyel- 
<!/ped  in  powers  of  the  number.     For,  if  possible,  suppose 

log  X  :=  Cq  +  C^x  +  Cc^x^  +  etc. 
Supposing  a;  =  0,  we  have 

Co  =  log  0, 
which  we  have  found  to  be  negative  infinity   (§  312,  VI). 
Hence  the  development  is  impossible. 

But  we  can  develop  log  (1  +  y)  in  powers  of  y.  For  this 
purpose,  we  develop  (1  +  yY  hy  both  the  binomial  and  expo- 
nential theorems,  and  compare  the  coefficients  of  the  first 
power  of  x.    First,  the  binomial  theorem  gives 

/-.    .     \^       -.                x(x  —  l)^      x(x—l)(x—'^)^^, 
(1  -^y)x^i  +  xy-\^       172"     y  +  "^ — 172^3 ^2/Hetc. 


382  LOGARITHMS. 

If  we  develop  the  coefficients  of  if,  y^,  etc.,  by  performing 
the  multiplications,  we  have 

Coef.  of  if  =  — f-i^ ;  part  in  :?;  r=  —  -. 

In  general,  in  the  coefficient  of  y^,  or 

x{x-^l)(x-2),.\.{x-n-\-l),  .^ 

the  term  containing  the  first  power  of  x  will  be 

±1>2»3 {n  —  l)x  _       X 

1.2.3 n  ~^  ^n 

Hence, 

(l  +  y)^  =il+x\^y  —  I  +  I  —  |-  +  etc.j  + terms  in  x\  a^,  etcJ 

On  the  other  hand,  the  exponential  development,  §  309,  (8), 
gives,  by  putting  1  +  ^  for  c^.  , 

(1  +  y)^  =  1  +  C^x  -\-  terms  in  x%  a^,  etc.  ^ 

Equating  the  coefficients  of  x  in  these  two  identical  series 
we  have 

^i=2/-f +  |'-^+etc.  (1) 

The  value  of  C^  is  given  by  the  theorem  of  §  310,  putting 
1  -^'y  ioY  a]  that  is,  C^  is  here  defined  by  the  equation 

6^-  =il  -\-  y. 

Hence,  if  we  take  the  number  e  (§  310)  as  the  base  of  a 
system  of  logarithms,  we  shall  have 

G,  =  log  (1  +  y). 

Comparing  with  (1),  we  reach  the  conclusion: 

Theorem.    Assuming  the  J^aperian  base  e  as  a  base, 
we  have 

log  (1  +  2/)  =  2/  -  f-  +  I  -  |-  +  etc.,  ad  inf.       (2) 


LOGARITHMS.  383 

Def.  Logarithms  to  the  base  e  are  called  Naperian 
Logarithms,  or  Natural  Logarithms. 

The  appellation  "  natural "  is  used,  because  this  is  the  simplest  system 
of  logarithms. 

Kem.  The  series  (2)  is  not  convergent  when  y  ">  1,  and 
therefore  must  be  transformed  for  use. 

Putting  —  y  for  y  in  (2),  we  have 

log(l--y)  =  -^-|  -|  -^tc- 
Subtracting  this  from  (2),  and  noticing  that 
log  (l+y)-  log  (1  -  2/)  =  log  ^  (Th.  VIII), 

we  have         log  ^^  =  2y +  ^ +  ^  +  etc.  (3) 

Now  n  being  any  number  of  which  we  wish  to  investigate 
ippose  y  =  ^_ 

1-y  n     ' 

whence     log  ip-^-  =  log =  log  (^^  +  1)  —  log  n. 

Substituting  these  values  in  (3),  we  have 

log  {n  +  1)  -log..  ^  ^  +  3^-iy3  +  ^p^-iy5 

+  etc.  (4) 

This  series  enables  us  to  find  log  {n  +  1)  when  we  know 
log  n.  To  find  log  2,  we  put  ^  =  1,  which,  because  log  1 
=  0,  gives 

log2  =  3g  +  3^3  +  A3  +  ^  +  etc.) 

Summing  five  terms  of  this  series,  we  find 
log  2  =.  0.693147 


the  logarithm,  let  us  suppose  y  =  - — — —  •    This  will  give 


384  L00ARITHM8, 

Putting  ^^  =r  2  in  (4),  we  have 

log3=:log2  +  2g  +  3l3  +  J^  +  J^  +  etc.),        ^ 

which  gives  log  3  —  1.098612. 

Because  9  =  32,      log  9  :=  2  log  3  =  2.197224. 
Putting  ^  =  9  in  (4),  we  have 

log  10  =  log  9  +  2  (1  +  3i-3  +  ^~  +  etc.), 

whence  log  10  =  2.302585. 

In  this  way  the  [N^aperian  logarithms  of  all  numbers  may  be 
computed.  It  is  only  necessary  to  compute  the  logarithms  of 
the  prime  numbers  from  the  series,  because  those  of  the  com- 
posite numbers  can  be  formed  by  adding  the  logarithms  of 
their  prime  factors.     (§  312,  VII.) 

314,  Definitive  Form  of  the  Exponential  Series,  We  are 
now  prepared  to  give  the  exponential  series  (§  309,  8)  its  defi- 
nite form.     Since  the  coefficient  0^  is  defined  by  the  equation 

e^i  =  a, 
the  quantity  G  is  the  Naperian  logarithm  of  a.     Hence,  the 
exponential  series  is 

«.  =  1  +  ^l2g^  +  (^«1  +  (^M^V  etc., 

which  is  a  fundamental  development  in  Algebra. 

By  putting  «  =  e,  we  have  log  a  =  1,  and  the  series  be- 
comes that  for  e^  already  found. 

By  putting  x=il,  we  have  an  expression  for  any  number 
in  terms  of  its  natural  logarithm,  namely, 


Comparison  of  Two  Systems  of  Logarithms. 

315.  Put        e,  the  base  of  one  system  ; 
a,  the  base  of  another; 
riy  a  number  of  which  we  take  the  logarithm 
in  both  systems. 


LOGARITHMS.  385 

Putting  I  and  V  for  the  logarithms  in  the  two  systems,  we 

have 

^  =1  n,  d'  =  n^ 

and  therefore  e^  =  d',  (1) 

Now  put  h  for  the  logarithm  of  a  to  the  base  e.     Then 

e^  —  a, 

and  raising  both  members  to  the  V*^  power, 

e^^'  =  d'. 

Comparing  with  (1),       I  =  kV, 

or  ^'  =  ^  X  i-  (2) 

This  equation  is  entirely  independent  of  7i,  and  is  therefore 
the  same  for  all  values  of  ii.     Hence, 

Theoeem.  //  ive  multiply  the  logarithm  of  any 
nuTThber  to  the  base  a  by  the  logarithin  of  a  to  the  base  e, 
we  shall  have  the  logarithin  of  the  number  to  the  base  e. 

316.  Although  there  may  be  any  nnmber  of  systems 
of  logarithms,  only  two  are  used  in  practice,  namely : 

1.  The  natural  or  Naperian  system,  base  =  ^  = 
2.718282 

2.  The  common  system,  base  — 10. 

The  natural  system  is  used  for  purely  algebraic 
purposes. 

The  common  system  is  used  to  facilitate  numerical 
calculations. 

Assigning  these  values  to  e  and  a  in  the  preceding  section, 
the  constant  k  is  the  natural  logarithm  of  10,  which  we  have 
found  to  be  2.302585. 

Therefore,  by  (2),  for  any  number, 

nat.  log  =  common  log  x  2.302585, 

,  ,  nat.  lo2r. 

and  common  log  -  ^^^^i;;;^ 

=:  nat.  log  X  0.4342944..,. 
Hence, 

25 


386  LOGARITHMS, 

Theoeem.  The  comnion  logarithm  of  any  ninnher 
vianj  he  found  by  multiplying  its  natural  logarithm  by 
0,4342944  ....  or  by  the  reciprocal  of  the  JS^aperian  loga- 
rithm of  10. 

Def,  The  number  0.4342944  is  called  the  Modulus 
of  the  common  system  of  logarithms. 

EXERCISES. 

1.  Show  that  if  a  and  h  be  any  two  bases,  the  logarithm  of 
a  to  the  base  b  and  the  logarithm  of  b  to  the  base  a  are  the  re- 
ciprocals of  each  other. 

2.  What  does  this  theorem  express  in  the  case  of  the  natu- 
ral and  common  systems  of  logarithms  ? 

Common  Logarithms. 
317.  Because 
102    --  100, 
101    ^    10^ 

100    ^      1^ 

_    1      )   we  have  to  base  10, 
^^     -   10^ 

10-2   _    __ 

^^      -  100' 

etc.  etc. 

The  following  conclusions  respecting  common  logarithms 
will  be  evident  from  an  inspection  of  the  above  examples : 

I.  The  logarithm  of  any  number  between  1  and  10  is 
a  fraction  between  0  and  1. 

II.  The  logarithm  of  a  number  ivith  two  digits  is  1 
plus  some  fraction. 

III.  In  general,  the  logarithm  of  a  number  of  i  digits 
is  i  —  1,  plus  some  fraction. 

IV.  The  logarithmj  of  a  fraction  less  than  unity  is 
negative. 

V.  Tlie  logarithms  of  two  numbers,  the  reciprocal  of 
each  other,  are  equal  and  of  opposite  signs. 


\ 


log 

100  = 

3, 

log 

10  = 

1, 

log 

1  = 

0, 

log 

1 

To  ~ 

-1, 

log 

1 

100  ~ 

-2, 

LOGARITHMS.  387 

VI.  //  one  number  is  10  times  another,  its  logarithm^ 
will  he  greater  by  unity. 

Proof.     If  10^  =  n, 

then  101+^  =  10  X  l(y  =  lOn, 

Hence,  if  I  =  log  n, 

then  ?  +  1  =  log  10^. 

318.  To  give  an  idea  of  the  progression  of  logarithms^  the 
following  table  of  logarithms  of  the  first  11  numbers  should  be 
studied. 

The  logarithms  are  not  exact,  because  all  logarithms,  ex- 
cept those  of  powers  of  10,  are  irrational  numbers,  and  there- 
fore when  expressed  as  decimals  extend  out  indefinitely.  We 
give  only  the  first  two  decimals. 

log  1  =  0.00,  log  7    =  0.85, 

log  2  =:  0.30,  logs    =:  0.90, 

log  3  =  0.48,  log  9    =  0.95, 

log  4  =  0.60,  log  10  =  1.00, 

log  5  =  0.70,  log  11  =  1.04. 
log  6  =r  0.78, 

It  will  be  noticed  that  the  difference  between  two  consecu- 
tive logarithms  continually  diminishes  as  the  numbers  increase. 
For  instance,  the  difference  between  log  20  and  log  10  must 
by  §  312,  YIII,  be  the  same  as  between  log  1  and  log  2. 

319.  Gomjmtation  of  Logarithms,  Since  the  logarithms 
of  all  composite  numbers  may  be  found  by  adding  the  loga- 
rithms of  their  factors,  it  is  only  necessary  to  show  how  the 
logarithms  of  prime  numbers  are  computed.  We  have  already 
shown  (§  313)  how  the  natural  logarithms  may  be  computed, 
and  (§  316)  how  the  common  ones  may  be  derived  from  them 
by  multiplying  by  the  modulus  0.4342944....  It  is  not  how- 
ever necessary  to  multiply  the  whole  logarithm  by  this  factor, 
but  we  may  proceed  thus: 

We  have,  putting  M  for  the  modulus, 

com.  log  n  =z  Mneii.  log  n, 

com.  log  {n  i-  1)  zzz  M  nat.  log  {n  -j-  1) ; 


388  L0GABITHM8, 

whence,  by  subtraction, 

com.  log  {n  + 1)  —  com.  log  n  =:^  M  [nat.  log  {n  -f  1) — nat.  log  n] ; 

and,  by  substituting    for     nat.  log  (^  +  1)  -—  nat.  log  n    its 
yalue,  §  313, 

com. log  {n  +  1)  =  com.  log  n  +  2m[-^^  +  372]^-VI? 

1 

+    T-JK ^Tr    +    etc. 

By  means  of  this  series,  the  computations  of  the  successive 
logarithms  may  be  carried  to  any  extent. 

Tables  of  the  logarithms  of  numbers  up  100,000,  to  seven  places  of 
decimals,  are  in  common  use  for  astronomical  and  mathematical  calcula- 
tions. One  table  to  ten  decimals  was  published  about  the  beginning  of 
the  present  century.  The  most  extended  tables  ever  undertaken  were 
constructed  under  the  auspices  of  the  French  government  about  1795,  and 
have  been  known  under  the  name  of  Les  Grandes  Tables  du  Cadadre. 
Many  of  the  logarithms  were  carried  to  nineteen  places  of  decimals. 
They  were  never  published,  but  are  preserved  in  manuscript. 

330.  It  may  interest  the  student  who  is  fond  of  computa- 
tion to  show  how  the  common  logarithms  of  small  numbers 
may  be  computed  by  a  method  based  immediately  on  first 
principles. 

Let  7Z  be  a  number,  and  let  -  be  an  approximate  value  of 
its  logarithm.     We  shall  then  have, 

n  =  10^, 
or,  raising  to  the  q^^  power, 

flQ  —  lOP. 

Hence,  could  we  find  a  power  of  the  number  which  is  also 
a  power  of  10,  the  ratio  of  the  exponents  would  at  once  give 
the  logarithm.  This  can  never  be  exactly  done  with  whole 
numbers,  but,  if  the  condition  be  approximately  fulfilled,  we 
shall  have  an  approximate  value  of  the  logarithm. 

Let  us  seek  log  2  in  this  way.  Forming  the  successive 
powers  of  2,  we  find 

210  _  1024  =  103  (1.024).  (1) 


LOGARITHMS.  389 

Hence,  3  :  10  =  0.3  is  an  approximation  to  log  2.  To 
find  a  second  approximation,  we  form  the  powers  of  1.024 
until  we  reach  a  number  nearly  equal  to  2  or  10,  or  the  quo- 
tient of  any  power  of  2  by  a  power  of  10.  Suppose,  for  instance, 
that  we  find 

1.024^^  =  2. 

Because  1.024  =  2^^-r- 10^,  this  equation  will  give 


'  2^^\^ 

=  )i,     or    Z''^^^  =  ^.lu 

dx 


)x 
=  2,     or    210^^  =  2. 103^,     or    2io^-i  =  lO^a^, 


which  will  give  log  2  = 


lOic- 


If  we  form  the  powers  of  1.024  by  the  binomial  theorem, 
or  in  any  other  way,  we  shall  find  that  x  is  between  29  and  30, 
from  which  we  conclude  that  log  2  =  0.301  nearly. 

To  obtain  a  yet  more  exact  value,  we  form  the  30th  power 
of  1.024  to  six  or  seven  decimals,  and  put  it  in  the  form 

1.02430  =  2  (1  +  a), 

where  a  will  be  a  small  fraction. 

Then  we  find  what  power  of  1  -\-  cc'  will  make  2.  Let  y  be 
this  power.  Kaising  the  last  equation  to  the  yth  power,  we 
have 

1.02430y  =  2^(1  +  a)y  =  2y+K 

Putting  for  1.024  its  value,  2^^  -r- 10^,  this  equation  becomes 

2^1/ 


10902/ 


2^+1,        or        2299y-i  =  1090y, 


whence,.  log  2  =  g^|^. 

By  a  little  care,  tlie  value  of  y  can  be  obtained  so  accurately 
that  the  value  of  log  2  shall  be  correct  to  8,  9,  or  10  places  of 
decimals. 

The  power  to  which  we  must  raise  1  +  cc  to  produce  2  will 

be  approximately  — — — ~ — ,  when  a  is  very  small. 


390  LOGARITHMS. 

EXERCISES. 

1.  In  the  common  system  {a  =  10)  we  have 

log  2  =  0.30103,  log  3  =z  0.47712. 

Hence  find  the  logarithms  of  4,  5,  6,  8,  9,  12,  12^,  15,  16, 
16|,  18,  20,  250,  6250. 

Note  that  5  =  ^ ,  12J  =  if ^,  16|  =  if^,  and  apply  Th.  VIII. 

2.  How  many  digits  are  there  in  the  hundredth  power  of  2? 

3.  Given  log  49  =  1.690196;  find  log  7. 

4.  Given  log  1331  =  3.124178 ;  find  log  11. 

5.  Find  the  logarithm  of  105  and  1.05  from  the  above  data  ? 

6.  Find  the  logarithm  of  1,06^^, 

7.  If  $1  is  put  out  at  5  per  cent,  per  annum  compound 
interest  for  1000  years,  how  many  digits  will  be  required  to 
express  the  amount?     (Compare  §  216.) 

8.  Prove  the  equation 

log  X  =  ^  log  (:?;  -f-  1)  +  ^  log  {x--l) 


+  M 


+  o  /o^9         '-1  \.q  +  K  /o^5!         iTs  +  ^^^' 


_2x^  _  1  ^  3  (2:r2  —  1)3  ^  5  {2x^  —  1)' 

9.  lty  =  log  71,  of  what  numbers  will  y  -i-  ly  y  +  2,  y  —  1 
and  «/  —  2  be  the  logarithms  ? 

10.  Find  X  from  the  equation  cP^  =:lu 

Solution.     Taking  the  logarithms  of  both  members,  we  have 
X  log  c  =  log  h ; 

whence,  x  =  :r^—  • 

logo 

1 

11.  c^  =  n.  12.     (^  =  —> 

m 

13.     b^  =  -.  14.     h-^  =  p. 

Show  that  the  answers  to  (13)  and  (14)  are  and  ought  to  be  identical. 
15.     a^^  z=z  m,  16.     bc^  =  h 

17.     Find  X  and  y  from  the  equations 


] 


BOOK    XII. 
IMA  GIN  A  RY     QUANTI  TIE  S. 


CHAPTER    1. 
OPERATIONS    WITH    THE    IMAGINARY    UNIT.* 

331.  Since  the  square  of  either  a  negative  or  a  positive 
quantity  is  always  positive,  it  follows  that  if  we  have  to  extract 
the  square  root  of  a  negative  quantity,  no  answer  is  possible, 
in  ordinary  positive  or  negative  numbers  (§§  170,  200). 

In  order  to  deal  with  such  cases,  mathematicians  have  been 
led  to  suppose  or  imagine  a  kind  of  numbers  of  which  the 
squares  shall  be  negative.  These  numbers  are  called  Imagi- 
nary Quantities,  and  their  units  are  called  Imaginary 
Units,  to  distinguish  them  from  the  ordinary  positive  and 
negative  quantities,  which  are  called  real. 

332,  The  Imaginary  Unit,  Let  us  have  to  extract  iJie 
square  root  of  —  9.  It  cannot  be  equal  to  +  3  nor  to  —  3, 
because  the  square  of  each  of  these  quantities  is  +  9.  We 
may  therefore  call  the  root  V—  9,  just  as  we  put  the  sign  \/ 
before  any  other  quantity  of  which  the  root  cannot  be  extracted. 
But  the  root  may  be  transformed  in  this  way  : 

Since  —  9  =  +9  x  —1, 

it  follows  from  §  183  that  • 

*  It  is  not  to  be  expected  that  a  beginner  will  fully  understand  tliis 
subject  at  once.  But  he  should  be  drilled  in  the  mechanical  process  of 
operating  with  imaginaries,  even  though  he  does  not  at  first  understand 
their  significance,  until  the  subject  becomes  clear  through  familiarity. 


392  IMAOmART   QUANTITIES. 

Def,  The  surd  V— 1  is  the  Imaginary  Unit.  Tlie 
imaginary  unit  is  commonly  expressed  by  the  symbol  1 

This  symbol  is  used  because  it  is  easier  to  write  i  than 

The  unit  Hs  a  supposed  quantity  such  that,  when  squared, 
the  result  is  —  1. 

That  is,  i  is  defined  by  the  equation 
%^  =  —  1. 

Theoeem.  Tl%e  square  root  of  any  negative  quantity 
may  he  expressed  as  a  number  of  imaginary  units. 

For  let  —  ^  be  the  number  of  which  the  root  is  required. 

Then  V  —  n  =z  V  -\-  n  V—  1  =  VnL 

Hence, 

To  extract  the  square  root  of  a  negative  quantity , 
extract  the  root  as  if  the  quantity  were  positive,  and 
affix  the  symbol  i  to  it. 

333.  Complex  Quantities,  In  ordinary  algebra,  any  num- 
ber may  be  supposed  to  mean  so  many  units.  7  or  a,  for 
example,  is  made  up  of  7  units  or  a  units,  and  might  be  writ- 
ten 7-1  or  al. 

When  we  introduce  imaginary  quantities,  we  consider  them 
as  made  up  of  a  certain  number  of  imaginary  units,  each  repre- 
sented by  the  sign  i,  just  as  the  real  unit  is  represented  by  the 
sign  1.     A  number  h  of  imaginary  units  is  therefore  written  hi, 

A  sum  of  a  real  units  and  h  imaginary  units  is  written 
a  +  hi, 
and  is  called  a  complex  quantity.     Hence, 

Def.  A  Complex  Quantity  consists  of  the  sum  of 
a  certain  number  of  real  units  plus  a  certain  number  of 
imaginary  units. 

Def.  When  any  expression  containing  the  symbol 
of  the  imaginary  unit  is  reduced  to  the  form  of  a  com- 
plex quantity,  it  is  said  to  be  expressed  in  its  Normal 
Form. 


IM AGIN  ART  QUANTITIES.  393 

Addition  of  Complex  Expressions. 

334.  The  algebraic  operations  of  addition  and  subtraction 
are  performed  on  imaginary  quantities  according  to  nearly  the 
same  rules  which  govern  the  casQ  of  surds  (§  181),  the  surd 
being  replaced  by  L     Thus, 

aV—  1  +  b^/—  I  =  ai  -\-  U  =  {a -\-  b)  i. 

Hence  the  following  rule  for  the  addition  and  subtraction 
of  imaginary  quantities : 

Add  or  subtract  all  the  real  terms,  as  in  ordinaj^j 
algebra.  Then  add  the  coefficients  of  the  imaginary 
unit,  and  affix  the  symbol  i  to  their  sum. 

Example.  Add  a  -t-  bi,  6  +  li,  5  —  10/,  and  subtract 
^a  —  2bi  +  z  from  the  sum. 

We  may  arrange  the  work  as  follows: 

a  +    bi 
6  4-    W 
5  —  lOi 
—  z  —  da  -{-  2bi    (sign  changed). 

Sum,      —z  —  2a-{-ll  +  {3b  —  3)  i. 

EXE  RC  I  S  E  S. 

1.  Add  dx  +  ^yi  +  m,  2m  +  6ni,  6m  —  62/i. 

2.  Add  4a^,  17 i,  3a  +  6bi,  x  +  yi. 

3.  From  the  sum  a  -{-  bi  -\-  m  —  ni  —  ;^  +  qi  subtra<)t  the 
sum  ■\-  yi  —  z  —  ui. 

Reduce  to  the  normal  form : 

4.  '  a  -{-bi  —  {m  —  ni)  —  {x  +  yi), 

5.  7n{a  —  bi)  —  n{x  —  yi). 

Multiplication  of  Complex  Quantities. 

325.  Theorem.  All  the  even  powers  of  the  imagi- 
nary unit  are  real  units,  and  all  its  odd  powers  are 
imaginary  units,  positive  or  negative. 


394  IM AGIN  ART   QUANTITIES, 

Proof.  The  imaginary  unit  is  by  definition  such  a  symbol 
as  when  squared  will  make  —  1.     Hence, 

^2  =  - 1. 

Now  multiply  both  sides  of  this  equation  by  i  a  number  of 
times  in  succession,  and  substitute  for  each  power  of  i  its  value      ■ 
given  by  the  preceding  equation.     We  then  have 

i^  =  —  ^2  =rr  -f  1  (because  v^  =.  —  1), 
i^  —  _«4  —  -f  ^2  —  —  1, 

etc.      etc.  etc. 

It  is  evident  that  the  successive  powers  of  i  will  always 
have  one  of  the  four  values,  %  —  1,   —  i,  or  +  1. 

iy  i^,  ^^,  etc.,  will  be  equal  to         i; 

t\  ^^  i^o,  etc.,          ''              "  —1; 

^,  P,  P-\  etc.,          ''              "  -  ^; 

^^  ^8,  ^l2^  etc.,          "              "  ■\-  1. 

We  may  express  this  result  thus :        > 
If  n  is  any  integer,  then: 

To  multiply  or  divide  imaginary  quantities,  we  proceed  as 
if  they  were  real  and  substitute  for  each  power  of  %  its  value  as 
a  real  or  imaginary,  positive  or  negative  unit. 

Ex.  I.     Multiply  ai  by  xi. 

By  the  ordinary  method,  we  should  have  the  product, 
axi'^.     But  i^  =  —  1.     The  product  is  therefore  —  ax. 

That  is,  ai  x  xi  =:  —  ax, 

Ex.  2.  Multiply  a  -f  bi  by  m  +  ?ii, 

ni  (a  +  hi)  =  ani  —  dn  (because  ni  x  M  =  —hn) 
m  {a  -f  M)  =  hmi  +  rt??z 
(77i  +  ni)  {a  +  ^i)  =  «77z  —  ^^^  +  {an  +  Z>w)  i, 
which  is  the  product  required. 


IMAOINART   QUANTITIES,  395 

EXERCISES 

Multiply 

I.  X  -\-  yi  hy  a  —  b.  2,  m  -\-  7ii  by  ai, 

3.  m  —  ni  by  M.  4.  1  +  ^  by  1  —  i. 

5.  a;  —  y^  by  ^  +  M.  6.  a;  —  «/*  hy  x  -\-  yi. 

7.  a  —  «^^  —  M  by  a  +  ai  +  M. 
Develop 

8.  (flj  +  My.  9.  (7?z  +  7^^)3. 
10.  (1  +  ^)2.  II.  (l  —  iy. 

336.  Imaginary  Factors.  The  introduction  of  imaginary 
units  enables  us  to  factor  expressions  which  are  prime  when 
only  real  factors  are  admitted.  The  following  are  the  princi- 
pal forms : 

a^^b^  =  {a  +  M)  {a  —  bi), 

a^-W±  Ubi  =  {a±  bif. 

The  first  form   shows  that  the  sum  of  two  squares  can 
always  be  expressed  as  a  product  of  two  complex  factors. 
For  example,  17  =  4^  +  12  =  (4  +  i)  (4  —  i). 

EXERCISES. 

Factor  the  expressions : 

I.     x^ -\- 4..  2.     x^-\-2. 

3.  ^^^^x  +  b  =1  (x  —  1)2  +  4. 

4.  x^  —  4:X  -}-  13.  5.     a  -\-  b. 

6.     a2  ^  2an  +  5^^2.  7.     x^  +  2xy  +  2y\ 

337.  Fu:n^damental  Prin'ciple.  A  complex  quantity 
A  -\-  Bi  cannot  he  equal  to  zero  unless  zve  have  hoth 

A  :=zO        and        B  =  0. 

Proof.  If  ^  and  ^  were  not  zero,  the  equation  A-\-Biz=zO 
would  give 

A 

that  is,  the  imaginary  unit  equal  to  a  real  fraction,  which  is 
impossible. 

Cor.     If  both  members  of  an  equation  containing  imagi- 


396  IMAGINARY   QUANTITIES. 

nary  units  are  reduced  to  the  normal  form,  so  that  the  equation 

shall  be  in  the  form 

A-\-Bi  =  M-^Ni, 

we  must  haye  the  two  equations, 

A  =  M,  B  =z  N, 

For,  by  transposition,  we  obtain 

A^M^{B-N)i  =  0, 

whence  the  theorem  giyes  A  —  M=iO^  B  —  JSF=0.     Hence, 

Every  equation  hetween  complejo  quantities  involves 
two  equations  hetween  real  quantities,  formed  by  equating 
the  numhers  of  real  and  imaginary  units. 

Reduction    of  Functions    of  i   to   the  Normal 

Form. 

338.    1.  If  we  have  an  entire  function  of  i, 

a  -\-M  -\-  ci^  +  di^  -f  ei^  -{-fi^  -f  etc., 
we  reduce  it  by  putting 

i^  =  —  1,     i^  =  —  i,     i^  =  1,    etc.,     etc., 
and  the  expression  will  become 

{a  —  c  -^  e  —  etc.)  +  {b  —  d  +/—  etc.)^; 
which,  when  we  put 

X  =  a —  c -\- e  —  etc,        y  =  h  —  d -i-f  —  etc, 
becomes  x  +  yi,  as  required. 

2.  To  reduce  a  rational  fraction  of  i  to  the  normal  form, 
we  reduce  both  numerator  and  denominator.  The  fraction 
will  then  take  the  form 

a  -\-  bi 
m  +  7ii 

Since  this  is  to  be  reduced  to  the  form  x  -f  yiy  let  us  put 

a  -^-bi 

— - — -.  =  X  -\-  yi, 

m  -}-  m  ^ 

X  and  y  being  indeterminate  coefficients. 

Clearing  of  fractions, 

a  -\-  bi  =  mx  —  7iy  +  (my  -f  nx)  i. 


IMA  OINAB  7    Q  UANTITIES,  397 

Comparing  the  number  of  real  and  imaginary  units  on 
each  side  of  the  equation,  we  have  the  two  equations 

7nx  —  ny^ztty  nx  +  my  =  5. 

Solving  them,  we  find 

ma  -\-  nb  _  mb  —  na 

*,,       ^  a  -\-  hi        ma  +  nh      ml)  —  na  . 

Therefore,     — ^ — .  =  — ^— — ^  -\ ^— — ^  t, 

771  4-  ni        m^  +  w-        7)1^  +  11? 

which  is  the  normal  form. 


EXERCISES. 

Eeduce  to  the  normal  form  : 


2.     1  +  i  —  i^  +  i^  —  i^  —  1*5  +  i\         '''     i^i 

6  +  5i  1  +  ^*  ^      ^2^*  (^  —  c^i) 

A,     — - — :•  q.      — — ;•  6.     ^^ -' 

^      (j  —  6i  ^      1  —  ^  X  -\-  ai 

1  —  i  a  -\-U  (a-\-M)  {a— hi) 

7-     2~T5'  •     'a^Yi  ^'  (^^i)2~-" 

10.  What  is  the  value  of  the  exponential  series  which  gives 
the  development  of  e^?    We  put  x  =  i  in  §  310,  Eq.  10. 

11.  Develop  (1  +  xiY  hj  the  binomial  theorem. 

12.  What  are  the  developed  values  of 

(1  +  UY  +  (1  —  uy 

and  (1  -i^biY  —  {l  —  UY^ 

13.  Write  eight  terms  of  the  geometrical  progression  of 
which  the  first  term  is  a  and  the  common  ratio  t. 

14.  Find  the  limit  of  the  sum  of  the  geometrical  progres- 

sion  of  which  the  first  term  is  a  and  the  common  ratio  -• 

z 

339.  To  reduce  the  square  root  of  an  imaginary  expres- 
sion to  the  normal  form. 

Let  the  square  root  be  ^a  +  bi. 


We  put  X  +  yi  z=  ^J a  +  bi. 

Squaring,      x^  —  y'^  -\-  %xyi  =  «5  +  bi. 


398  IM AGIN  ART   QUANTITIES. 

Comparing  units,      x^  -—  ?/  =  a, 

Solving  this  pair  of  quadratic  equations,  we  find 
_  V{Va^-\-^  +  a) 
V2  ' 

\/(V^3+  I)' -a) 


Therefore, 


'=  V2 


EXERCISES. 

Eeduce  the  square  roots  of  the  following  expressions  to  the 
normal  form : 

I.     3  +  4^.  2.     4  +  di.  3.     12  +  5^. 

4.  Find  the  square  roots  of  the  imaginary  unit  i,  and 
of  —  i,  and  prove  the  results  by  squaring  them. 

Note  that  this  comes  under  the  preceding  fomi  when  a  =  0  and 
6  =  ±1. 

5.  Find  the  fourth  roots  of  the  same  quantities  by  extract- 
ing the  square  roots  of  these  roots. 

330.  Quadratic  Equations  loith  Imaginary  Roots,  The 
combination  of  the  preceding  operations  will  enable  us  to  solve 
any  quadratic  equation,  whether  it  does  or  does  not  contain 
imaginary  quantities. 

Example  i.    Find  x  from  the  equation 

x^  -^  43:  +  13  =  0. 
Completing  the  square  and  proceeding  as  usual,  we  find 

x^  -^  4cx-^  4.  =  —  9, 

whence  a;  +  2  =  a/—  9  —  ±3/, 

and  X  =z  _  2  ±  3^. 

Ex.  2.  x^  +  Ixi  —  c  =  0. 

Completing  the  square, 

X^  +  OXl  —  —   :=  C  —  —' 

4  4 


IMAGINABT  QUANTITIES,  399 

Extracting  the  root, 

hi  _  VJg^^g 

^■^  2  "         2        ' 

1  hi 

whence  x  =  ±,-  a/(4c  —  h^)  —  — • 

EXERCISES. 

Solve  the  quadratic  equations: 

I.     a;2  -f-  a;  +  1  =  0.  2.     2;2  —  a;  +  1  —  0. 

3.     a;2  +  3^2;  -f  10  r=  0.  4.     :zj2  _|_  iqo;  +  34  =  0. 

Form  quadratic  equations  (§  199)  of  which  the  roots  shall  be 
5.     a  +  hi  and  a  —  hi.  6.     ai  +  h  and  ai  —  h, 

331.  Exponential  Functions.  When  in  the  exponential 
function  a^  we  suppose  z  to  represent  an  imaginary  expression 
X  +  «/i,  it  becomes 

This  expression  could  have  no  meaning  in  any  of  our  pre- 
vious definitions  of  an  exponent,  because  we  have  not  shown 
what  an  imaginary  exponent  could  mean.  But  if  we  suppose 
the  effect  of  the  exponent  to  be  defined  by  the  exponential 
theorem  (§§  309,  314),  we  can  develop  the  above  expression. 
First  we  have,  by  the  fundamental  law  of  exponents, 
f^x+yi  —  a^avK 

Next,  if  we  put  c  =  Nap.  log  a,  we  have 
a  —  e<^; 
whence,  ay^  =  ePVK 

K  we  put,  for  brevity,  cy  =  u,  we  shall  now  have 

The  value  of  a^  being  already  perfectly  understood,  we 
may  leave  it  out  of  consideration  for  the  present,  and  investi- 
gate the  development  of  e^*.    By  the  exponential  theorem 

(§  310,  10), 

.       ^    .      .      ii^ij^      u^i?      tiH^      uH^ 
e«i  =  1  +  m  +  -p  +  -3  J  +  -J-  +  -J-  +  etc. 


400  IMAQINABT   QUANTITIES. 

Substituting  for  the  powers  of  i  their  values  (§  325), 

.»  =  1  -  -J  +  jj  -  gj  +  etc.  +  ^M  -  ^  +  gj  -  etc.)  I. 

These  two  series  are  each  functions  of  u,  to  which  special 
names  have  been  given,  namely : 

Def.  The  series  1  —  ^,  +  ji  —  ^.  +  51  —  etc.,  is  called 
the  cosine  of  if,  and  is  written  cos  u, 

Def.    The  series  '^  —  07+^7  —  7"^  +  ^^  —  ^^c-?   is  called 

the  sine  of  u^  and  is  written  sin  u. 

Using  this  notation,  the  above  development  becomes, 

QUI  —  COS  u  +  i  sin  u,  {a) 

which  is  a  fundamental  equation  of  Algebra,  and  should  be 
memorized. 

Remarks.  These  functions,  qo&  u  and  sin  u,  have  an  ex- 
tensive use  in  both  Trigonometry  and  Algebra.  To  familiarize 
himself  with  them,  it  will  be  well  for  the  student  to  compute 
their  values  from  the  above  series  for  u  =  0.25,  u  =  0.50, 
u=zl,  u=:2,  to  three  or  four  places  of  decimals.  This  can 
be  done  by  a  process  similar  to  that  employed  in  computing  e 
in  §  310.    If  the  work  is  done  correctly,  he  will  find : 

For    u  =  ~, 
4 

"   -  =  l' 

"        U  =zl, 

"      u  =  2, 

333.  Let  us  now  investigate  the  properties  of  the  functions 
cos  u  and  sin  u,  which  are  defined  by  the  equations, 

u^      ii'^      u^ 
cos^.  =  l-^  +  jj--^  +  etc. 

^         ^         7  /  (^) 

U^         VP         w  ,       ^ 

miu  —  u  —  —  Ar  -^^—  -^  +  etc. 


1 

COS  -7  = 

4 

0.969, 

sin  \  =.  0.247 
4 

cos|  = 

0.878, 

sin  1  =  0.479. 

COS  1   = 

0.540, 

sin  1  =  0.841 

COS  2  = 

-  0.416, 

sin  2  —  0.909. 

IM AGIN  ART   QUANTITIES,  401 

Since  cos  ti  includes  only  even  powers  of  u,  its  value  will 
remain  unchanged  when  we  change  the  sign  of  u  from  +  to 
— 5  or  vice  versa.    Hence, 

cos  {^u)  =z  cos  ti.  (1) 

Since  sin  u  contains  only  odd  powers  of  u,  its  sign  will 
change  with  that  of  u.     Hence, 

sin  (—  -w)  =  —  sin  u,  (2) 

If  in  the  equation  {a)  we  change  the  sign  of  Uy  we  have, 
by  (1)  and  (2), 

or  e""^*  =  cos  u  —  I  sin  u, 

Now  multiply  this  equation  by  {a).     Since 

1 

^m  X  e-'^^  =:  e^*  X  -rr  =  1, 

we  have  1  =  (cos  ?^)^  —  i^  (sin  -2^)2, 

or  1  =  (cos  uj^  +  (sin  uy. 

It  is  customary  to  write  cos^  u  and  sin^  u  instead  of  (cos  uf 
and  (sin  uY,  to  express  the  square  of  the  cosine  and  of  the 
sine  of  u.     The  last  equation  will  then  be  written 

cos^  u  +  sin^  u  ==  1.  {c) 

Although  we  have  deduced  this  equation  with  entire  rigor, 
it  will  be  interesting  to  test  it  by  squaring  the  equations  {V), 
First  squaring  cos  u,  we  find  (§  284), 


C0S2  ^  =  1  -^  ,,2  +  ,,4  (i.    ^  _i__  +  _)  « 


etc. 


The  coefficient  of  u"^  is  found  to  be 

1        1 1^ 1 

n\  +2!  (?z  — 2)!  "^4!(7^— 4)!  "^  ^  7i\ 

when  ^  is  double  an  even  number,  and  to  the  negative  of  this 
expression  when  n  is  double  an  odd  number. 
Again,  taking  the  square  of  sin  u,  we  find 


sm'*  u 
26 


^.  +  ,,4(__l^j^-J-^--)  +  etc. 


.402  IMAGINARY   QUANTITIES. 

the  coefficient  of  u'^  being 

1  1  1 


1 !  0*  —  1) !      3 !  (/i  —  3) !      5 !  (^  —  5) ! 


(^-1)!  1!' 

or  the  negative  of  this  expression,  according  as  ;^  ^  is  even  or 
odd.  ^ 

Adding  sin^  i^  and  cos^  u,  we  see  that  the  terms  u^  cancel 
each  other,  and  that  the  sum  of  the  coefficients  of  w*  can  be 
arranged  in  the  form 

^ 1_       _1 1_        1 

4!       1!  3!  "^2!  2!      3!  l!  "^4!* 

Let  us  call  this  sum  A,  If  we  multiply  all  the  terms  by 
4 ! ,  and  note  that  by  the  general  form  of  the  binomial  coeffi- 
cients, 

n\  _  ln\ 

wefind         4!^  =  l^g)+g)-(3^)+g), 

which  sum  is  zero,  by  §  2G2,  Th.  11.     Therefore  the  coefficients 
of  u^  cancel  each  other. 

Taking  the  sum  of  the  coefficients  of  u'^,  we  arrange  them 
in  the  form 

n\      1!  0^-1)!  "^2!  (?^-2)!      3!  (^^  -  3)!  +  ^  ^'^ 
which  call  A,     Then  multiplying  by  n\,  we  have 

»^^=i-©+(i)-(i)+--+e). 

which  sum  is  zero.  Therefore  all  the  coefficients  of  u'^  cancel 
each  other  in  the  sum  sin^  u  +  cos^  u,  leaving  only  the  first 
term  1  in  cos^  u,  thus  proving  the  equation  (6)  independently. 
This  example  illustrates  the  consistency  which  pervades  all 
branches  of  mathematics  when  the  reasoning  is  correct.  The 
conclusion  (c)  was  reached  by  a  very  long  process,  resting  on 
many  of  the  fundamental  principles  of  Algebra  ;  and  on  reach- 


IMAOINAUY  QUANTITIES.  403 

ing  a  simple  conclusion  of  this  kind  in  such  a  way,  the  mathe- 
matician always  likes  to  test  its  correctness  by  a  direct  process, 
when  possible. 

Let  us  now  resume  the  fundamental  equation  {a).  Since 
u  may  here  be  any  quantity  whatever,  let  us  put  nu  for  u. 
The  equation  then  becomes, 

^nui  _  cos  nu  +  i  sin  nu. 

But  by  raising  the  equation  {a)  to  the  n*^  power,  we  have 

Hence  we  have  the  remarkable  relation, 

(cos  u  -\-  i  sin  u)^  :=  cos  nu  +  i  sin  nu. 

Supposing  n=z2,  and  developing  the  first  member,  we 
have 

cos^  u  —  sin^  u  +  2i  sin  u  cos  tc  =z  cos  2u  -f  i  sin  2u. 
Equating  the  real  and  imaginary  parts  (§327,  Cor.),  we  have 
cos^  u  —  sin2  to  =  cos  2u, 
2  sin  u  cos  i^  =  sin  2u, 

relations  which  can  be  verified  from  the  series  representing 
cos  u  and  sin  u,  in  a  way  similar  to  that  by  which  we  verified 
sin^  2c  +  cos^  u  =  1. 

EXERCISES. 

1.  Find  the  values  of  cos^  u,  sin^  u,  cos*  u,  and  sin*  u  by 
the  preceding  process. 

2.  Write  the  three  equations  which  we  obtain  by  putting 
u  =.  a,  u  =z  b,  and  u  =i  a  -\-  h  in  equation  [a).  Then  equate 
the  product  of  the  first  two  to  the  third,  and  show  that 

cos  {a  -{-  h)  =.  cos  a  cos  h  —  sin  a  sin  Z>, 
sin  (a  4-  ^)  =  sin  a  cos  h  +  cos  a  sin  ^. 

3.  Eeduce  to  the  normal  form, 

{x  —  i)  {x  —  2i)  {x  —  30  {x  —  U). 

4.  Develop  (a^  +  U)^  by  the  binomial  theorem,  and  reduci 
the  result  to  the  normal  form. " 


404  QEOMETBIG  REPRESENTATION. 


CHAPTER     II. 

THE    GEOMETRIC     REPRESENTATION     OF    IMAGINARY 
QUANTITIES. 

333.  In  Algebra  and  allied  branches  of  the  higher  raaUie- 
maties^  the  fundamental  operations  of  Arithmetic  are  extended 
and  generalized.  In  Elementary  Algebra  we  have  already  had 
several  instances  of  this  extension,  and  as  we  are  now  to  have 
a  much  wider  extension  of  the  operations  of  addition  and  mul- 
tiplication, attention  should  be  directed  to  the  principles 
involved. 

In  the  beginning  of  Algebra,  we  have  seen  the  operation  of 
addition,  which  in  Arithmetic  necessarily  implies  increase,  so 
used  as  to  produce  diminution. 

The  reason  of  this  is  that  Arithmetic  does  not  recognize 
negative  quantities  as  Algebra  does,  and  therefore  in  employ- 
ing the  latter  we  have  to  extend  the  meaning  of  addition,  so  as 
to  apply  it  to  negative  quantities.  When  thus  applied,  we 
have  seen  that  it  should  mean  to  subtract  the  quantity  which 
is  negative. 

In  its  primitive  sense,  as  used  in  the  third  operation  of 
Arithmetic,  the  word  JnuUiplij  means  to  add  a  quantity  to  itself 
a  certain  number  of  times.  In  this  sense,  there  would  be  no 
meaning  to  the  words  "  multiply  by  a  fraction.^'  But  we  ex- 
tend the  meaning  of  the  word  multiply  to  this  case  by  defining 
it  to  mean  taking  a  fraction  of  the  quantity  to  be  multiplied. 
We  then  find  that  the  rules  of  multiplication  will  all  apply  to 
this  extended  operation. 

This  extension  of  multiplication  to  fractions  does  not  take 
account  of  negative  multipliers.  In  the  latter  case  we  can 
extend  the  meaning  of  the  operation  by  providing  that  the 
algebraic  sign  of  the  quantity  shall  be  changed  when  the  mul- 
tiplier is  negative.  We  thus  have  a  result  for  multiplication 
by  every  positive  or  negative  algebraic  number. 

Now  that  we  have  to  use  imaginary  quantities  as  multi- 


OEO METRIC  REPRESENTATION.  405 

pliers,  a  still  further  extension  is  necessary.  Hitherto  our 
operations  with  imaginary  units  have  been  purely  symbolic ; 
that  is,  we  have  used  our  symbols  and  performed  our  operations 
without  assigning  any  definite  meaning  to  them.  We  shall 
now  assign  a  geometric  signification  to  operations  with  imagi- 
nary units,  subject  to  these  three  necessary  conditions  : 

1.  The  operations  must  be  subject  to  the  same  rules  as 
those  of  real  quantities. 

2.  The  result  of  operating  with  an  imaginary  quantity 
must  be  totally  different  from  that  of  operating  with  a  real  one, 
and  the  imaginary  quantity  must  signify  something  which  a 
real  quantity  does  not  take  account  of. 

3.  If  the  imaginary  quantity  changes  into  a  real  one,  the 
operation  must  change  into  the  corresponding  one  with  real 
quantities. 

334.  Geometric  Represenfatmi  of  Imaginary  Units.  Cer- 
tain propositions  respecting  the  geometric  representation  of 
multiplication  have  been  fully  elucidated  in  Part  I,  and  are 
now  repeated,  to  introduce  the  corresponding  representations 
of  complex  quantities. 

I.  All  real  numbers,  positive  and  negative,  may  be  arranged 
along  a  line,  the  positive  numbers  increasing  in  one  direction, 
the  negative  ones  in  the  opposite  direction  from  a  fixed  zero 
point.  Any  number  may  then  be  represented  in  magnitude 
by  a  line  extending  from  0  to  the  place  it  occupies. 

We  call  this  line  a  Vector. 

II.  If  a  number  a  be  multiplied  by  a  positive  multiplier 
(for  simplicity,  suppose  +1),  the  direction  of  its  vector  will 
remain  unaltered.  If  it  be  multiplied  by  a  negative  multiplier 
(suppose  — 1),  its  vector  will  be  turned  in  the  opposite  direc- 
tion (from  0  —  a  to  0  -\-  a,  or  vice  versa).  Compare  §  72, 
where  the  coarse  lines  are  the  vectors  of  the  several  quantities. 

—a  0  +« 

1 I I 


III.  If  the  number  be  multiphed  twice  by  —  1,  that  is,  by 
(—1)2,  its  vector  will  be  restored  to  its  first  position,  being 
twice  turned,  and  if  it  be  multiplied  twice  by  +  1,  that  is,  by 
(-[-  1)%  its  vector  will  not  be  changed  at  all.     Its  vector  will 


406 


IMAOINABT  QUANTITIES. 


+ia 


therefore  be  found  in  its  first  position,  whether  we  multiply  it 
by  the  square  of  a  positive  or  of  a  negative  unit;  in  other 
words,  both  squares  are  positive. 

IV.  To  multiply  the  hne  +  a  twice  by  the  imaginary  unit 
i,  is  the  same  as  multiplying  it  by  i^  or  —  1.     Hence, 

Multiplying  by  the  iiivaginary  unit  i  must  give  the 
vector  such  a  motion  as,  if  repeated,  will  change  it  from 
-\-  a  to  —  a. 

Such  a  motion  is  given  by  turn- 
ing the  vector  through  a  right  angle, 
into  the  position  +  ia,     A  second 

motion   brings  it  to   the   position       __^ 

—  a,  the  opposite  of  -h  «.  A  third 
motion  brings  it  to  —  ia,  a  position 
the  opposite  of  -f  ia,  A  fourth 
motion  restores  it  to  the  original 
position  +  a. 

If  we  call  each  of  these  motions  multiplying  hy  i,  we  have. 


-ia 


la,  i^a 


a,  i^a 


ta. 


_4  -3  —2  —1 

! \ [ L_ 


+4i 


from  the  diagram,  a  =  a,  ia 

ikt  =  a,  which  corresponds  exactly  to  the  law  governing  the 

powers  of  i  (§  325).     Hence : 

//  a  quantity  is  represented  hy  a  vector  extending 
from  a  zero  point,  the  multiplication  of  this  quantity  hy 
the  imaginary  unit  may  he  represented  hy  turning  the 
vector  through  90°. 

V.  In  order  that  multiplier 
and  multiplicand  may  in  this  op- 
eration be  interchanged  without 
affecting  the  product,  we  must 
suppose  that  the  vertical  line 
which  we  have  called  ia  is  the 
same  as  ai,  that  is,  that  this  line 
represents  a  imaginary  units. 

We  have  therefore  to  count 
the  imaginary  units  along  a 

vertical  line  on  the  saine  system  that  we  count  the  real 
units  on  a  horizontal  line. 


+  i 


1      2 


^-2i 

3i 

U 


OEOMETRIG   HEP  RESENT ATION. 


407 


'«+6i 


U 


U 


7>^ 


bi 


\ 


^bi 


835.  Qeoinetric  Representation  of  a  Complex  Quantity, 
We  have  shown  (§  15)  that  algebraic  addition  may  be  represented 
by  putting  lines  end  to  end,  the 
zero  point  of  each  line  added  be- 
ing at  the  end  of  the  line  next 
preceding.  The  distance  of  the 
end  of  the  last  line  from  the  zero 
point  is  the  algebraic  sum. 

On  the  same  system,  to  repre- 
sent the  algebraic  sum  of  the  real 
and  imaginary  quantities  a  +  Ji, 
we  lay  off  a  units  on  the  real  (horizontal)  line,  and  then  h 
units  from  the  end  of  this  line  in  a  vertical  direction.  The 
-  end  of  the  vertical  line  will  then  be  the  position  corresponding 
to  flj  -f  hi. 

It  is  evident  that  we  should  reach  the  same  point  if  we 
first  laid  off  h  units  from  0  on  the  imaginary  line,  and  then  a 
units  horizontally.     Hence  this  system  gives 

})%  j^  a  =^  a  -\-  hi, 
as  it  ought  to,  to  represent  addition. 

If  «  or  J  is  negative,  it  is  to  be  laid  off  in  the  opposite  di- 
rection from  the  positive  one.  We  then  have  the  points  cor- 
responding to  —  a  -\-l)i,  —  a  —  hi,  and  aj  —  hi,  shown  in  the 
diagram,  which  should  be  carefully  studied  by  the  pupil. 

The  result  we  have  reached  is  the  following : 

Every  complex  quantity  a  -f  hi  is  considered  as  be- 
longing to  a  certain  point  on  the  plane,  namely,  that 
point  which  is  reached  by  laying  off  from  the  zero  point 
a  units  in  the  horizontal  direction  and  h  units  in  the 
vertical  direction, 

R 

336.  Addition  of  Com- 
plex Quantities,  If  we  have 
several  complex  terms  to 
add,  as  a  +  hi,  m  —  ni, 
p  +  qi,  we  may  lay  them 
off  separately  in  their  ap- 
propriate magnitude  and  di- 


408  IMAGINABT  QUANTITIES, 

rection,    as  in   the   figure,   the   last  line    terminating    in    a 
point  K. 

If  we  first  add  the  quantities  a  -\-  U,  etc.,  algebraically 
(§  324),  the  result  will  be 

a  +  m  -\-  p  -\-  {b  —  n  -\-  q)i. 

We  may  lay  off  this  sum  in  one  operation.  The  sum  a-hm- 
-\-p  will  carry  us  from  0  to  M,  and  the  sum  {b  ^  n  -\-  q)  i 
from  M  to  R,  because  MU=b  —  n  +  q.  Therefore  we  shall 
reach  the  same  point  R  whether  we  lay  the  quantities  off  sepa- 
rately, or  take  their  sum  and  lay  off  its  real  and  imaginary 
parts  separately. 

337.  Vectors  of  Complex  Quantities,  The  question  now 
arises  by  what  straight  line  or  vector  shall  we  represent  a  sum 
of  complex  quantities  ?     The  answer  is : 

T]^e  vector  of  a  sum  of  sev- 
eral vectors  is  the  straight  line 
from  the  heginning  of  the  first 
to  the  end  of  the  last  vector 
added. 

For  example,  the  sum  of  the 
quantities  OX  ==  a  and  XP  =  bi  is  the  vector  OP. 

It  might  seem  to  the  student  that  the  length  of  the  vector  represent- 
ing the  sum  should  be  equal  to  the  combined  lengths  of  all  the  separate 
vectors.  This  difficulty  is  of  the  same  kind  as  that  encountered  by  the 
beginner  in  finding  the  sum  of  a  positive  and  negative  quantity  less  than 
either  of  them.  The  solution  of  the  difficulty  is  simply  that  by  addition 
we  now  mean  something  different  from  both  arithmetical  and  algebraic 
addition.  But  the  operation  reduces  to  arithmetical  addition  when  the 
quantities  are  all  real  and  positive,  because  the  vectors  are  then  all  placed 
end  to  end  in  the  same  straight  line.  Therefore  there  is  no  inconsistency 
between  the  two  operations. 

Two  imaginary  quantities  are  not  equal,  unless  both  their 
real  and  imaginary  parts  are  equal,  so  that  their  sum  shall  ter- 
minate at  the  same  point  P.  Their  vectors  will  then  coincide 
with  each  other.    Hence : 

Tivo  vectors  are  not  considered  equal  unless  they  agree 
in  direction  as  well  as  length. 


GEOMETRIC  REPRESENTATION,  409 

In  other  words,  in  order  to  determine  a  vector  com- 
pletely, we  must  know  its  direction  as  well  as  its  length. 

This  result  embodies  the  theorem  of  the  preceding  chapter  (§  327), 
that  two  complex  quantities  are  not  equal  unless  both  their  real  and 
imaginary  parts  are  equal.  It  is  only  in  case  of  this  double  equality  that 
the  two  complex  quantities  will  belong  to  the  same  point  on  the  plane. 

Because  OXP  is  a  right  angle,  we  have  by  the  Pythagorean 
theorem  of  Geometry, 

(length  of  veetor)2  =  a^  _|_  j2^ 


or  length  of  vector  =  ^/d^  -j-  b\ 

We  are  careful  to  say  length  of  vector,  and  not  merely  vec- 
tor, because  the  vector  has  direction  as  well  as  length,  and  the 
direction  is  as  important  an  element  as  length. 

To  avoid  repeating  the  words  "  length  of,"  we  shall  put  a 
dash  over  the  letters  representing  a  vector  when  we  consider 
only  its  length.     Then  OX  will  mean  length  of  the  line  OX. 

Def.  The  length  of  the  vector,  or  the  expression 
Va^  -f  b%  is  called  the  Modulus  of  the  complex  ex- 
pression a  4-  M. 

The  modulus  is  the  absolute  value  of  the  expression,  con- 
sidered without  respect  to  its  being  positive  or  negative,  real 
or  imaginary.     Thus  the  different  expressions, 

—  5,     +5,     3  +  4/,    4  —  3i,    5^, 

all  have  the  modulus  5  (because  Vs^  4-  4^  =:  5).  The  points 
which  represent  them  are  all  5  units  distant  from  the  zero 
point,  and  so  lie  on  a  circle,  and  their  vectors  are  all  5  units  in 
length. 

The  German  mathematicians  therefore  call  the  modulus 
the  absohite  value  of  the  complex  quantity,  and  this  is  really 
a  better  term  than  the  English  expression  modulus, 

Def.  The  Angle  of  the  vector  is  the  angle  which  it 
makes  with  the  line  along  which  the  real  units  are 
measured. 

If  OA  is  this  line,  and  OB  the  vector,  the  angle  is  AOB. 


2. 

4  -  31 

3- 

—  4  +  Si. 

5- 

3  +  4^. 

6. 

3  —  4^. 

8. 

—   3    —  4:1. 

9- 

5  +  7i. 

II. 

5  +  6i, 

12. 

5  +  4i. 

14. 

3  +  i- 

15. 

S^i. 

nta" 

I  and  vertical 

line 

;    mark   several 

410  IM AGIN  ART  QUANTITIES. 

EXERCISES. 

Lay  off  the  following  complex  quantities,  draw  the  vectors 
corresponding  to  them,  and  find  the  modulus  both  by  measure- 
ment and  calculation  : 

I.  4  +  U. 

4.  —  4  —  3^. 

7.  —  3  -f  U. 

10.  5  +  6i. 

13.  3  +  2i. 

16.  3  —  2^. 

17.  Draw  a  he 

points  on  the  plane  of  these  lines,  and  find  by  measurement 
the  complex  expressions  for  each  point.  Also,  draw  the  sev- 
eral vectors  and  measure  their  length.  Continue  this  exercise 
until  the  relation  between  the  complex  expressions  and  their 

•points  is  well  apprehended. 

I^OTE.  The  student  may  adopt  any  scale  he  pleases,  but  a 
scale  of  millimeters  will  be  found  convenient. 

338.  Geometric  Multiplication.  The  question  next  arises 
whether  the  results  we  obtain  for  multiplication  of  complex 
quantities  follow,  in  all  respects,  the  usual  laws  of  multiplica- 
tion, especially  the  commutative  and  distributive  laws. 

I.  To  multiply  a  vector  hy  a  real  factor. 

Let  the  vector  he  a  -\-  hi  and  the  jf 

factor  m.     The  product  will  be  ^    ,-^'1 

ma  +  mhi. 

In  the  geometric  construction,  let 
OK  =  a    and  Al^^U.     We    shall     O- 
then  have,  by  the  rule  of  addition. 

Vector  OB  =:  a  +  U. 

"When  we  multiply  a  by  m,  let  OA'  be  the  product  ma,  and 
A'B'  the  product  mhL  Because  the  lines  OA  and  AB  are  both 
multiplied  by  the  same  real  factor  m  to  form  OA'  and  A'B',  we 
shall  have 

OA  :  AB  :  OB  =  OA'  :  A'B'  :  OB'. 


GEOMETRIC  REPRESENTATION.  411 

Therefore  the  triangles  OAB  and  OA'B'  are  similar  and 
equiangular,  so  that 

angle  A'OB'  =  angle  AOB. 

This  shows  that  the  lines  OB  and  OB'  coincide,  so  that 
BB'  is  the  continuation  of  OB  in  the  same  straight  line.  More- 
over, the  above  proportion  gives 

OB'  =  7??0B, 
or,  from  (1),         vector  OB'  =  m  vector  OB. 

Therefore,  multiplying  a  vector  hy  a  real  factor 
changes  its  length  without  altering  its  direction. 

II.  To  multiply  a  vector  hy  the  imaginary  unit. 

Multiplying  a  +  hi  by  i,  the        ^q 
result  is 

—  b  +  ai. 

The   construction  of  the  two     '^' 
vectors  being  made  as  in  the  fig- 
ure, we  have  p  _y 
OB  =  a  +  M, 
OQ.  =  —  /^  +  ai. 

Because  the  triangles  OPQ  and  OAB  are  right-angled  at  P 
and  B,  and  have  the  sides  containing  the  right  angle  equal  in 
length,  they  are  identically  equal,  and 

angle  POQ  =  angle  OBA  =z  90^  —  angle  BOA. 
Hence  the  sum  of  the  angles  POQ  and  BOA  is  a  right 
angle,  and  because  POA  is  a  straight  line,  therefore, 
angle  BOQ  .-=  90°. 

Therefore,  the  result  of  multiplying  the  vector  OB  by 
the  imaginary  unit  is  to  turn  it  90°  ivithout  changing 
its  length. 

We  have  assumed  this  to  be  the  case  when  the  vector  represents  a 
real  quantity,  or  lies  along  the  line  OB  ;  we  now  see  that  the  same  thing 
holds  true  when  the  vector  represents  a  complex  quantity. 

If  instead  of  the  multiplier  being  simply  the  imaginary 
unit,  it  is  of  the  form  ni,  then,  by  (I),  in  addition  to  turning 
the  vector  through  90°,  we  multiply  it  by  n. 


412 


IMAGINARY   QUANTITIES, 


III.  To  multiply  a  vector  by  a  complex  quantity, 
m  +  ni. 

This  will  consist  in  multiplying  separately  by  m  and  ni^ 
and  adding  the  two  products.     Put  OB  =  a  +  6^,  the  yector 
to  be  multiplied ;  ON  = 
m  +  ni,  the  multiplier. 

To  multiply  OB  by  m^ 
we  take  a  length  OC,  deter- 
mined by  the  proportion, 

OC  :  OB  =  m  :  1,    (I) 

whence  by  (I), 

OC  =  m-OB 
=  w  (a  +  hi). 

To  multiply  OB  by  ni,  we  take  a  length  CD  determined 
by  the  condition, 

length  GY)  —  n  length  OB, 

CD  :  OB 


or 


nil; 


and  to  multiply  by  i,  we  place  it  perpendicular  to  OB.      (II) 
We  then  have, 

CD  =  OB  X  ni. 

In  order  to  add  it  to  OC,  the  other  product,  we  place  it  as 
in  the  diagram,  and  thus  find  a  point  D  which  corresponds  to 
the  sum 

OC  +  CD  =  OBxm  +  OBxm'; 
that  is,  to  the  product 

{m  +  ni)  (a  +  hi). 
Now  because  OC  ==  OB  x  m  and  CD  =  OB  x  n,  we  have 
OC"  :  CD  =  m:n=zOM  :  MN, 

and  because  the  angles  at  M  and  C  are  right  angles,  the  tri- 
angles OCD  and  OMN  are  similar.     Therefore, 

angle  COD  =  angle  MON. 

Hence  the  angle  AOD  of  the  product- vector  is  equal  to  the 
sum  of  the  angles  of  the  multiplier  and  multiplicand. 
For  the  length  OD  of  the  product-vector  we  have, 


GEOMETBIG   REPRESENTATION^,  413 

length  0&  =  00^  +  CJ? 

=  (m2  +  n^)  OBl 
Extracting  the  square  root, 

length  OD  =      Vni^+n^  •  OB 

Therefore  the  length  of  the  product-vector  is  equal  to  the 
products  of  the  lengths  of  the  vectors  of  the  factors. 

Combining  these  two  results,  we  reach  the  conclusion: 

The  modulus  of  the  produet  of  two  coinplex  factors  is 
equal  to  the  product  of  their  moduli. 

The  angle  of  the  product  is  equal  to  the  sum;  of  the 
angles  of  the  factors. 

339.    TJie  Roots  of  Unity.     We  f? 

have  the  following  curious  problem :  |  Ny> 

Given,  a  vector  0 A,  which  call  « ;  i     \ 

it  is  required  to  find  a  complex  factor  i        \ 

X,  such  that  when  we  multiply  a   n  ¥ 70  A 

times  by  x,  the  last  product  shall  be  a  / 

itself.     That  is,  we  must  have  / 


o^a  =  a. 


/C 


The  required  factor  must  be  one 
which  will  turn  the  vector  round  without  changing  its  length. 
Let  us  begin  with  the  case  of  n=:3. 

Since  three  equal  motions  must  restore  OA  to  its  original 
position,  the  condition  will  be  satisfied  by  letting  x  indicate  a 
motion  through  120°,  so  that  OA  shall  take  the  position  OB 
when  angle  AOB  =  120°.  Then,  P  being  the  foot  of  the  per- 
pendicular from  B  upon  AO  produced,  we  shall  have  angle 
POB  =  60°,  and  angle  PBO  =  30°.    Therefore, 

FO  =  la,  PB  =  ^^a, 

and  vector  OB  =:  xa  =z  —  -a  +  --;-aL 


414  IMAOINART  QUANTITIES. 

Because  the  factor  x  has  not  changed  the  length  of  the  line, 
the  modulus  of  x  is  unity,  and  because  it  has  turned  the  line 
through  120°,  its  angle  is  120°.     Therefore  its  value  is 

-OP-fPBi 

on  a  scale  of  numbers  in  which  OB  =  1 ;  that  is, 

1    ,    a/3. 

^==--2 +-2-^- 

Eeasoning  in  the  same  way  with  respect  to  the  product  x^a, 
which,  produces  the  vector  OC,  we  find 

^  -  ~  2  ~    2   '^ 

an  equation  which  we  readily  prove  by  squaring  the  preceding 
value  of  X  and  reducing. 

Multiplying  these  values  of  a:  and  x^,  v/e  find 

x^  =  1, 

which  ought  to  be  the  case,  because  x^a  =  a.     Hence, 

1        a/S 
-     The  complex  quantity  —  ^  H — \y-  i  ^s  a  cube  root  of 

unity. 

But  the  vector  00,  of  which  the  angle  is  240°,  also  repre- 
sents a  cube  root  of  unity,  if  we  suppose  00  =  1,  because 
three  motions  of  240°  each  turn  a  vector  through  720°,  or  two 
revolutions,  and  thus  restore  it  to  its  original  position.  This 
also  agrees  with  the  algebraic  process,  because,  by  squaring  the 
above  value  of  x^,  we  have 

^^       ^       -^       a/3.  1    .    a/3 


2   ^=-2  +  -2-^  =  ^^ 


V2         2    7~4       4+^ 
and  by  repeating  the  process  we  find 

Since  1  itself  is  a  cube  root  of  unity,  because  1^  =  1,  we 
conclude :  • 

There  are  three  cube  roots  of  unity. 


GEOMETRIC  REPRESENTATION.  415 

We  readily  find,  by  the  process  of  §  334,  IV,  that 

i,     —  1,     —  i,     and     1, 

are  all  fourth  roots  of  unity. 

By  a  course  of  reasoning  similar  to  the  above  for  any  value 
of  n,  we  conclude  : 

The  n^^  roots  of  unity  are  7i  in  number. 

EXERCISES. 

1.  Form  the  first  eight  powers  of  the  expression 

V2  +  V3   ' 

show  that  the  eighth  power  is  1,  and  lay  off  the  vector  corre- 
sponding to  each  power. 

2.  Form  the  first  twelve  powers  of 

a/3       1. 

and  show  that,  the  twelfth  power  is  +1. 

3.  Find  the  fifth  and  sixth  roots  of  unity  by  dividing  the  cir- 
cle into  five  and  six  parts,  and  either  computing  or  measuring 
the  lengths  of  the  Hnes  which  determine  the  expression. 

Note.  The  student  will  remark  the  similarity  of  the  gen- 
eral problem  of  the  n*^  roots  of  unity  to  that  of  dividing  the 
circle  into  n  equal  parts  (Geom.,  Book  VI). 


BOOK    XIII. 

777^    GENERAL     THEORY    OF  EQUA^ 
TIONS. 


Every  Equation  has  a  Root. 

340.  In  Book  III,  equations  containing  one  unknown 
quantity  were  reduced  to  the  normal  form 

Ax''  +  Bx"^-^  +  Gx^-'^  +  .  .  .  .  +  i^  =  0. 

If  we  divide  all  the  terms  of  this  equation  by  the  coefficient 
A^  and  put,  for  brevity, 

B 
Pi  =  J. 

C 

etc.      etc. 
F 

Pn  =  3, 

the  equation  will  become 

X^  +  p^X^-^  +i?2^"^  + -i-Pn-l^  -{-Pn  =  0.        (a) 

This  equation  is  called  the  General  Equation  of  the 
fith  Degree,  because  it  is  the  form  to  which  every  algebraic 
equation  can  be  reduced  by  assigning  the  proper  values  to  n, 
and  to  j^i,  P2,  Ps7  etc. 

The  n  quantities  Pi,  P29  *  »  »  *  Pn  are  called  the  Coeffi- 
cients of  the  equation. 

We  may  consider  pn  as  the  coefficient  of  xf^  =  1. 

341.  Theoeem  I.  Every  algebraic  equation  has  a  root, 
real  or  imaginary. 

That  is,  whatever  numbers  we  may  put  in  place  o^  p^,  P2, 
^3,  .  .  .  .  Pn,  there  is  always  some  expression,  real  or  imaginary, 
which,  being  substituted  for  x  in  the  equation,  will  satisfy  it. 


GENERAL    THEORY   OF  EQUATIONS.  ^Yl 

Rem.  The  theorem  that  every  equation  has  a  root  is  demonstrated  in 
special  treatises  on  the  theory  of  equations,  but  the  demonstration  is  too 
long  to  be  inserted  here. 

If  we  suppose  the  values  of  the  coefficients  p^p^}  etc.,  to 
vary,  the  roots  will  vary  also.    Hence, 

Theorem  II.  ITie  roots  of  an  algebraic  equation  are 
functions  of  its  coefficients. 

Example.  In  Chapter  YI  we  have  shown  that  the  roots 
of  a  quadratic  equation  are  functions  of  the  coefficients,  because 
if  the  equation  is 

x^  +  px  -^  q  ^=  0, 

the  root  is  x  =  — - — -— -, 

which  is  a  function  of  p  and  q. 

342,  Equations  tvliich  can  de  solved.  If  the  degree  of  the 
equation  is  not  higher  than  the  fourth,  it  is  always  possible  to 
express  the  root  algebraically  as  a  function  of  the  coefficients. 

But  if  the  equation  is  of  the  fifth  or  any  higher  degree,  it 
is  not  possible  to  express  the  value  of  the  root  of  the  general 
equation  by  any  algebraic  formulae  whatever. 

This  important  theorem  was  first  demonstrated  by  Abel  in 
1825.  Previous  to  that  time,  mathematicians  frequently  at- 
tempted to  solve  the  general  equation  of  the  fifth  degree,  but 
of  course  never  succeeded. 

This  restriction  applies  only  to  the  general  equation,  in 
which  the  coefficients  p^,  p^,  p^,  etc.,  are  all  represented  by 
separate  algebraic  symbols.  Such  special  values  may  be 
assigned  to  these  coefficients  that  equations  of  any  degree  shall 
be  soluble. 

343.  The  problem  of  finding  a  root  of  an  equation  of  the 
higher  degrees  is  generally  a  very  complex  one.  If,  however, 
the  equation  has  the  roots  —  1,  0,  or  +  1,  they  can  easily  be 
discovered  by  the  following  rules : 

I.  If  the  algebraic  sum  of  the  coefficients  in  the  equa- 
tion vanishes,  then  -\-lis  a  root. 

27 


418  GENERAL    THEORY   OF  EQUATIONS. 

II.  //  the  sum  of  the  coeffieients  of  the  even  powers  of 
X  is  equal  to  that  of  the  eoeffieients  of  the  odd  powers, 
then  —  1  is  a  root 

III.  //  the  absolute  term  p^  is  wanting,  then  0  is  a 
root. 

These  rules  are  readily  proved  by  putting  «=  +1,  then  x—  —1, 
then  a;  —  0  in  the  general  equation  {a)  and  noticing  what  it  then  reduces 
to.     The  demonstration  of  II  will  be  a  good  exercise  for  the  student. 

Number  of  Roots  of  General  Equation. 

344.  In  the  equation  (a),  the  left-hand  number  is  an  en- 
tire function  of  x,  which  is  equal  to  zero  when  the  equation  is 
satisfied.  Instead  of  supposing  an  equation,  let  us  suppose  x 
to  be  a  variable  quantity,  which  may  have  any  value  whatever, 
and  let  us  study  the  function  of  x, 

^  -{-p^x!>^-^+PzX^-^  + +Pn^lOC  +pn, 

which  for  brevity  we  may  call  Fx, 

Whatever  value  we  assign  to  x,  there  will  be  a  correspond- 
ing value  of  Fx, 

Example.    Consider  the  expression 

Fx  =  x^  —  7^2  -I-  36. 

Let  us  suppose  x  to  have  in  succession  the  values  —  4, 

—  3,-2,    —  1,   0,  1,  2,  etc.,  and  let  us  compute  the  corre- 
sponding values  of  Fx,    We  thus  find, 

X  =  —     4,     —    3,     —  2,     —    1,  0, 

Fx=z  —  140,     —  54,         0,     +  28,     +  36, 

xz=z         1,  2,    3,  4,  5,     6,  7,  S.'^ 

Fx=  -\-  30,     + 16,    0,     - 12,     — 14,    0,     +  36,     + 100. 

We  see  that  while  x  varies  from  —  4  to  +8,  the  value  of 
Fx  fluctuates,  being  first  negative,  then  changing  to  positive, 
then  back  to  negative  again,  and  finally  becoming  positive  once 
more. 

We  also  see  that  there  are  three  special  values  of  x,  namely, 

—  2,   +3,  and  +  6,  which  satisfy  the  equation  Fx  =  0,  and 
which  are  therefore  roots  of  this  equation. 


GENERAL    THEORY   OF  EQUATIONS, 


419 


345.  Representation  of  Fx  ly  a  Curve.  In  Book  VIII  ifc 
was  shown  how  a  function  of  a  variable  of  the  first  degree  might 
be  represented  to  the  eye  by  a  straight  line.  The  relation 
between  a  variable  and  any  function  of  it  may  be  represented 
to  the  eye  in  the  same  way  by  a  curve,  as  shown  in  Geometry, 
Book  VIL  We  take  a  base  line,  mark  a  zero  point  upon  it, 
and  lay  off  any  number  of  equidistant  values  of  x.  At  each 
point  we  erect  a  perpendicular  proportional  to  the  corresponding 
value  of  Fx  at  that  point,  and  draw  a  curve  through  the  ends. 


The  fluctuations  of  the  vertical  ordinates 
of  the  curve  now  show  to  the  eye  the  corre- 
sponding fluctuations  of  Fx, 

When  Fx  is  negative,  the  curve  is  below 

the  base  line.     When  Fx  is  positive,  the  curve 

is  above  the  base  line. 

The  roots  of  the  equation  Fx^=.^  are  shown  by  the  points 

at  which  the  curve  crosses  the  base  line.     In  the  present  case 

these  points  are  —  2*,    4-3,   -f-  6. 

In  order  to  distinguish  the  roots  from  the  variable  quantity 
Xj  we  may  call  them  «,  (3,  y,  d,  etc.,  or  x^^  x^,  x^,  etc.,  or  a^, 
^2,  ^3,  etc.,  the  symbol  x  being  reserved  for  the  variable. 

The  distinction  between  x  and  the  roots  will  then  be  this: 
X  is  an  independent  variable,  which  may  have  any  value 

whatever. 

Fx  is  a  function  of  x  of  which  the  value  is  fixed  by  that  of  x, 
a,  ft  y,  etc.,  or  a;  1,  a; 2,  3^3,  etc.,  are  special  values  of  x  which, 

being  substituted  for  x,  satisfy  the  equation 

Fx  =  0. 

Theorem.  An  equation'  mith  real  coefficients,  of  which 
the  degree  is  an  odd  number,  must  have  at  least  one  real 
root. 


Fx 


430  GENERAL    THEORY  OF  EQUATIONS. 

Proof.  1.  When  n  is  odd,  x'^  will  have  the  same  sign  (4- 
or  — )  as  X. 

2.  So  large  a  value,  positive  or  negative,  may  be  assigned  to 
X  that  the  term  x'^  shall  be  greater  in  absolute  magnitude  than 
all  the  other  terms  of  the  expression  Fx,  For,  let  us  put  the 
expression  Fx  in  the  form 

=  ^(i+^+5+....+|).       1) 

If  we  suppose  x  to  increase  indefinitely  either  in  the  posi- 
tive or  negative  direction,  the  terms    — ,  ^,  etc.,  will  all 
°  X      x^ 

approach  0  as  their  limit  (§  303,  Th.  I).    Therefore  the  expression 

\  ^-tl  ^(^  j^  etc.  will  approach  unity  as  its  limit,  and  will 

therefore  be  positive  for  large  values  of  x,  both  positive  and 
negative.  The  whole  expression  will  then  have  the  same  sign 
as  the  factor  x'^,  and,  n  being  odd,  will  have  the  same  sign  as  x, 

3.  Therefore,  between  the  value  of  x  for  which  Fx  is  negative 
and  that  for  which  it  is  positive  there  must  be  some  value  of  x 
for  which  Fx  =  0,  that  is,  some  root  of  the  equation  Fx  =  0. 

For  illustration,  take  the  preceding  cubic  equation. 

Cor.  An  equation  of  odd  degree  has  an  odd  numher 
of  real  roots. 

For,  as  Fx  changes  from  negative  to  positive  infinity,  it 
must  cross  zero  an  odd  number  of  times. 

346.  Theorem  I.  //  we  divide  the  expression  Fx  by 
X  —  a,  the  remainder  will  he  Fa,  or 

Kemainder  =  a^  -^  p^a^~^  +  p^oP''^  +  .  .  .  .  +  ptf 

Special  Illustration,     Let  the  student  divide 

^-^bx^  -{-  dx  H-  1 

by  x  —  a,  according  to  the  method  of  §  96.  He  will  find  the 
remainder  to  come  out 

a^-^ba^  -\-Za  +  1. 


GENERAL    THEORY   OF  EQUATIONS.  421 

General  Proof,     When  we  divide  Fxhy  x^  a,  let  us  put 
Qy  the  quotient ; 
R,  the  remainder. 

Then,  because  the  dividend  is  equal  to  the  product,  Divi- 
sor X  Quotient  +  Remainder, 

{x  —  a)Q  +  R=  Fx. 

Two  things  are  here  supposed: 

1.  That  this  equation  is  an  identical  one,  true  for  all  values 
of  X,  This  must  be  true,  because  we  have  made  no  supposition 
respecting  the  value  of  x, 

2,  That  we  have  carried  the  division  so  far  that  the  remain- 
der R  does  not  contain  x. 

Because  it  is  true  for  all  values  of  x,  it  will  remain  true 
when  we  put  x  =  a  on  both  sides.     It  thus  reduces  to 

R  =  F{a), 
which  is  the  theorem  enunciated. 

The  value  of  x  being  still  unrestricted,  let  us  in  dividing 
take  for  a  a  root  a  of  the  general  equation  Fx  =  0.  Then, 
by  supposing  x  =  a,  the  equation  (a)  will  be  satisfied,  or 

Fa  z=z  0. 

Therefore  if  we  divide  the  general  expression  Fx  hy  x -—  a, 
the  remainder  Fa  will  be  zero.     Hence. 

Theorem  II.  If  we  denote  by  a  a  root  of  the  eqitation 
Fx  =  0,  the  expression  Fx  will  be  exaetly  divisible  by 
X—  a. 

Illustration.     One  root  of  the  equation 
a;3  _  ^2  _  11^  _|.  15  _.  0 

is  3.    If.  we  divide  the  expression 

7^  —  x^  —  ll:r  +  15 
by  X  —  3,  we  shall  find  the  remainder  to  be  zero. 

347.  When  we  divide  Fxhj  x  —  a,  the  highest  power  of 
X  in  the  quotient  will  be  x^~K  Therefore  the  quotient  will  be 
an  entire  function  of  x  of  the  degree  n  —  1. 


422  GENERAL    THEORY   OF   EQUATIONS, 

Illustration,     The  quotient  from  the  last  division  was 
x^  +  2x-  5, 
which  is  of  the  second  degree,  while  the  original  expression  was  of  the 
third  degree. 

If  we  call  this  quotient  F^x,  we  shall  have,  by  multiplying 
divisor  and  quotient, 

Fx  =z  (x  -—  a)  F^x, 

Now  suppose  (i  a  root  of  the  equation 

i^i«  =  0 ; 

then  F^x  will,  by  the  preceding  theorem,  be  exactly  divisible 
hj  x  —  (i. 

The  quotient  from  this  division  will  be  an  entire  function 
of  X  of  the  degree  n  —  2.  This  function  may  again  be  divided 
by  x  —  y,  representing  by  y  the  root  of  the  equation  obtained 
by  putting  the  function  equal  to  zero,  and  so  on. 

The  results  of  these  successive  divisions  may  therefore  be 
expressed  in  the  form 

Fx  :=:  {x  —  a)  F^x  ....  (Degree  ^  —  1),  ^ 

F^x  —  {x  —  P)  F^x (Degree  ti  —  2),  >-  (1) 

F^x  =  {x  —  y)  F^x  ....  (Degree  n  —  d),) 
etc,  etc.      etc. 

Since  the  degree  is  diminished  by  unity  with  every  division, 
we  shall  at  length  have  a  quotient  of  the  first  degree  in  x,  of 

the  form 

x-^e,  ^ 

e  being  a  constant.  I 

Then,  by  substituting  in  the  equations  (1)  for  each  func- 
tion of  x  its  value  in  the  equation  next  below,  we  shall  have      ^ 

I 
Fx  =  (x  —  a){x  —  P){x-'y) {x --  e), 

the  number  of  factors  being  equal  to  the  degree  of  the  original 
equation.     Hence, 

Theorem  I.  Every  entire  function  of  x  of  the  nth 
degree  may  he  divided  into  n  factors,  each  of  the  first 
decree  in  x. 


NVMBER    OF  ROOTS.  423 

Since  a  product  of  several  factors  becomes  zero  whenever 
any  of  the  factors  is  zero,  it  follows  that  the  equation 

Fx  —  0 
will  be  satisfied  by  putting  x  equal  to  any  one  of  the  quantities 
€c,  P,  y,  ,  .  »  .  e,  because  in  either  case  the  product 

{x  —  a)(x  —  P)  [x  —  y)  .  ,  ,  ,  {x  —  e) 

will  vanish.     Therefore  the  quantities 

«,  /3,  y, £, 

are  all  roots  of  the  original  equation  Fx  =  0.     Hence, 

Theorem  IL  An  algebraic  equation  of  the  n^^  degree 
has  n  roots. 

We  have  seen  (§  195)  that  a  quadratic  equation  has  two 
roots.  In  the  same  way,  a  cubic  equation  has  three  roots,  one 
of  the  fourth  degree  four  roots,  etc. 

Moreover,  a  product  cannot  vanish  unless  one  of  the  factors 
vanishes.     Hence  the  product 

Fx     or     (x  —  a)(x  ^P){x  —  y^  ,  .  ,  ,  (x^  e) 

cannot  vanish  unless  x  is  equal  to  some  one  of  the  quantities, 
«,  i3,  y,  .  .  .  .  e.     Hence, 

An  equation  of  the  rfi^  degree  can  have  no  more  than 
n  roots. 

348.  We  may  form  an  equation  of  which  the  roots  shall 
be  any  given  quantities,  a,  Z>,  c,  etc.,  by  forming  the  product, 

(x  —  a)(x  —  h)  {x  —  c),  etc. 
Example.    Form  an  equation  of  which  the  roots  shall  be 

—  1,     -f  1,     1  +  2i,    1  —  2/. 
Solution,     We  form  the  product 

{x+l)(x^l){x-^l-  2i)  {x-^l  +  20, 
which  we  find  to  be 

x^  —  2a^  -i- ^x^  -{-  2x -- 5. 
Therefore  the  required  equation  is 

x^  —  2:i^-\-  4t2  -^2x  —  6  =  0. 


424  GENERAL    THEORY   OF  EQUATIONS. 

EXERCISES. 

Form  equations  with  the  roots : 

1.  2  +  V3,     2  ~  \/3,     —  2,     +1. 

2.  3  +  V5,     3  -  ^5,     -  3. 

3.  2,     -2,     4  +  V7,    4-V7. 

4.  1  +  V3,     1  -  V3,     1  +  V5,     1  -  V5. 

349.  When  we  can  find  one  root  of  an  equation,  then,  hy 
dividing  the  equation  by  x  minus  that  root,  we  shall  have  an 
equation  of  lower  degree,  the  roots  of  which  will  be  the  remain- 
ing roots  of  the  given  equation. 

Example.     One  root  of  the  equation 

x^  —  x^  —  11.T  +  15  =:  0 
is  3.     Find  the  other  two  roots. 

Dividing  the  given  equation  by  a:  —  3,  the  quotient  is 

x^  j^^x  —  5. 

Equating  this  to  zero,  we  have  a  quadratic  equation  of 
which  the  roots  are 

—  1  4-  V6     and     —  1  —  a/6. 

Hence  the  three  roots  of  the  original  equation  are 

3,     —  l+VO,     —  1  — V6. 

EXERCISES. 

1.  One  root  of  the  equation 

x^  _  32;2  —  Ux  +  12  =  0 
is— 3.     Find  the  other  two  roots. 

2.  Find  the  five  roots  of  the  equation 

a:5  _  4^  _^  I2x^  ^  4^2  __  13^  _-  0. 

(Compare  §  343.) 

350.  Equal  Roots,  Sometimes,  in  solving  an  equation, 
several  of  the  roots  may  be  identical. 

For  example,  the  equation 

2;3  _  6^2  _|.  12^  _  8  =  0 


COEFFICIENTS  AND    ROOTS.  425 

has  no  root  except  2.  If  we  divide  it  by  x  —  2,  and  solve  the 
resulting  quadratic,  its  roots  will  also  be  2.  Hence,  when  we 
factor  it  the  result  is 

{x  -2)(x  —  2)  (x-2)  =  0. 

In  this  case  the  equation  is  said  to  have  three  equal  roots. 
Hence,  in  general, 

TJte  n  roots  of  an  equation  of  the  n^^  degree  are  not  all 
necessarily  different  from  each  other,  hut  two  or  more  of 
them  may  he  equal. 

Relations  between  Coefficients  and  Roots. 

351.  Let  us  suppose  the  roots  of  the  general  equation  of 
the  n^^  degree 

x^  +  p^x^-^  +  2h^~^  +  .,..+  pn-\  ic  +  jt?»  =  0 
to  be  «,  ^,  y, .  .  .  .  €. 

We  have  shown  (§  341)  that  these  roots  are  functions  of 
the  coefficients  p-^,  p,^,  .  .  .  .  pn-.  To  find  these  functions  is  to 
solve  the  equation,  which  is  generally  a  very  difficult  problem. 

But  the  coefficients  can  also  be  expressed  as  functions  of 
the  roots,  and  this  is  a  very  simple  process  which  we  have 
already  performed  in  some  special  cases  by  forming  equations 
having  given  roots  (§  348). 

If  we  form  an  equation  with  the  two  roots,  a  and  i3,  the 
result  will  be 

.      0  =  {x  —  a){x-^(i)  =  x^—{a  +  p)x  +  af). 
Comparing  this  with  the  general  form, 
x^  -{-  p^x  4-i?2  =  ^> 
we  see  that  p^  =  —  [a  ^  (3)^ 

P2  =  ^P, 
a  result  already  reached  (§§  198,  199). 

Next  form  an  equation  with  the  three  roots,  a,  P,  y. 
Multiplying  {x  —  a)  (x  —  (3)  by  x  —  y,  we  find  the  equa- 
tion to  be 

a^—  {a  +  13  +  y)x^  +  {ap  -\-  Py  +  ya)x  —  aPy  =  0. 


426  GENERAL    THEORY   OF  EQUATIONS. 

So  in  this  case,    p^  =  —  («  +  /^  +  y), 
;?2  =  «3  +  Py  +  ya, 

Vz  =  —  «/^r- 

Adding  another  root  (5,  we  find  the  result  to  be 

Pi  =  -  («  +  ^  +  y  +  ^), 

;?2  =  «/3  +  c«y  +  «(5  +  i3y  +  /3(y  +  yd,  (2) 

Ps  =  —  ccpy  —  apd  —  ayd  —  jSyd, 
p^  =  €cpy6. 

Generalizing  this  process,  we  reach  the  following  conclu- 
sions : 

The  coefficient  p^  of  the  second  term  of  the  general  equa- 
tion is  equal  to  the  sum  of  the  roots  taken  negatively. 

The  coefficient  i^g  of  the  third  term  is  equal  to  the  sum  of 
the  products  of  every  combination  of  two  roots. 

The  coefficient  p^  of  the  fourth  term  is  equal  to  the  sum 
of  the  products  of  every  combination  of  three  roots  taken 
negatively. 

The  last  term  is  equal  to  the  continued  product  of  the  neg- 
atives of  the  roots. 

353.  Symmetric  Functions,  It  will  be  remarked  that  the 
preceding  expressions  for  the  coefficients  p^,  pc^,  etc.,  are  all 
symmetric  functions  of  the  roots  a,  p,  y,  etc,     (§  256.) 

The  following  more  extended  theorem  is  true  : 

Theorem.  Every  rational  symmetric  function  of  the 
roots  of  an  equation  may  he  expressed  as  a  rational 
function  of  the  coefficients. 

Example.     From  the  equations  (2)  we  find 

p,^  -  2p^  =  C.2  4.  /32  +  ^2  +  ^^ 
^PiP2  -Pi^  _  3j93  =  «3  +  i33  +  y3  +  ^. 

We  thus  reach  the  curious  conclusion  that  although  we 
may  not  be  able  to  find  any  individual  root  of  an  equation,  yet 
there  is  no  difficulty  in  finding  the  continued  product  of  the 
roots,  their  sum,  the  sum  of  their  squares,  of  their  cubes,  etc. 

The  general  demonstration  of  this  theorem,  and  the  methods  by  which 
any  rational  symmetrical  function  of  the  roots  may  be  determined,  are 
found  in  more  advanced  treatises. 


DERIVED   FUNCTIONS.  427 

Derived  Functions. 

353.  Def.    If  in  the  expression 

we  substitute  x^Ti  for  x^  and  then  develop  in  powers 
of  Ti^  the  coefficient  of  the  first  power  of  Ti  is  called  the 
First  Derived  Function  of  oc. 

To  find  the  First  Derived  Function.    Putting  x  -{-h  for  x, 
the  result  is 
F{x-\-Ji)  =  {x+h)^-{-p^{x  +  h)^-^  +  ,...+pn-i{^+Ji)+Pn'    (a) 

Developing  the  several  terms  of  the  second  member  by  the 
binomial  theorem,  we  have 

n  in  ——  1 1 

{x  +  hy  =  ^'^  +  nx^'-^n  H ^-- -x^-'^W  +  etc., 

{x  +  Tif-^  =  x"^-^  +  {n  —  1)  x'^-^h  +  etc., 
{x  +  ny-'^  =  x^-^  +  (n  —  'Z)  x^-^h  H-  etc., 
etc.  etc.  etc. 

Substituting  these  expressions  in  the  equation  {a)  and 
leaving  out  the  terms  in  h^,  h%  etc.  (because  we  do  not  want 
them),  we  have 

F{x  +  h)  =  x^  +  p^x'^-^  +  p^^x^''^  + +  Pn-i  ^  +  Pn 

+  [nx^'-^^{7i—l)  p^x'^-^-]-{n—'Z)p^x^-^-\- -\-pn-i\  h 

+  omitted  terms  7nultiplied  ly  li^,  W,  etc,  (b) 

We  see  that  the  first  line  is  here  the  original  Fx,  while  the 
coefficient  of  A  in  the  second  line  is  by  definition  the  derived 
function.     So,  if  we  put 

F'x,  the  derived  function  of  Fx, 
we  have     F{x  +  h)  =  Fx  +  7i  F'x  +  terms  x  h^,  h%  etc. 

Let  the  student,  as  an  exercise,  now  find  the  derived  function  of 
^4  +  30^3  _  5:^2  +  7^.  _  9 

by  the  process  just  followed,  commencing  with  equation  (a). 

Examining  the  coefficient  of  h  in  (Z>),  we  see  that  the  de- 
rived function  is  formed  by  the  following  rule  : 


428  GENERAL    THEORY   OF  EQUATIONS. 

Multiply  ecwli  term  by  the  exponent  of  the  variable  in 
that  term,  and  diminish  the  exponent  by  unity. 

The  last  or  constant  term  disappears  entirely  from  the  ex- 
pression. 

EXERCISES. 

Form  the  derived  function  of  the  following  expressions  : 

1.  ^5  _!_  5^  _l_  8^  _  2a;2  _  ^.  _l_  1. 

Ans.  bx^  +  20.^•3  +  24:^:2  _  4^.  __  1, 

2.  x^  —  2x^  —  2.^3  _  %x. 

3.  i^  +  12:^:5  _  242;3  -f  x^  +  7. 

4.  r^  —  2ax^  +  3¥x^  +  a^x. 

5.  x^  —  ^mx^  +  lOma^  —  16m.x\ 

Rem.  The  student  shoald  obtain  the  result  by  substituting  x-^h  toT 
X  in  each  equation  and  developing,  until  he  is  master  of  the  process. 

3o4,  Second  Form  of  the  Derived  Fmiction,  If,  as  be- 
fore, we  put  a,  f3,  y,  6,  etc.,  for  the  roots  of  the  equation 
Fx  =z  0,  we  shall  have 

Fx  =  {x  —  a)  (x  —  P)  {x  —  y)  .  ,  .  ,  {x  —  e).  (c) 

Let  us  form  the  derived  function  from  this  expression. 
Putting  X  -\-  h  for  x,  it  will  become 
{h  -i-  X  —  a)  {h  +  X  —  p)  {h  +  X  —  y) {h  -^  x  —  e). 

Studying  this  expression,  and  forming  the  products  which 
contain  h  when  three  or  four  factors  only  are  included,  we  see 
that  the  coefficient  of  the  h  in  the  first  factor  is  (x—p)  {x — y) 
. .  . . ,  in  the  second  factor  (x—a)  (x—y) . . . . ,  etc.  That  is, 
the  total  coefficient  of  h  will  be 

(x  —  p)  {x  —  y)  .  .  .  ,  {x  —  e),  omitting  first  term ; 
^  i^x  —  cc)  (x  —  y)  .  .  .  .  {x  —  e),  omitting  second  term ; 

etc.        etc.  etc. 

+  {x  —  a)  (x  —  P)  {x  —  y)  .  .  .  ,    omitting  last  term. 

But  comparing  with  (c),  we  see  that  the   first  of  these 

Fx  Fx 

products  is  ,  the  second  is  — — ^ ,  etc.,    to  the  last, 

X  —  cc  X  —  p 

Fx 
which  is Hence, 

X  —  e 


DERIVED   FUNCTIONS,  429 

X— ax  —  [3x  —  y  X  —  E       ^  ' 

Illustration,     Let  us  take  once  more  the  expression  of 

§  344, 

Fxz=zQ^^nx^  +  36, 

of  which  the  three  roots  are  —2/3,  and  6.    Its  derived  func- 
tion, by  method  (1),  is 

Zx^  —  Ux. 

Expressing  Fx  as  a  product  of  factors,  it  is 
Fx  =  {x  -i-  2)  {x  —  3)  {x  —  6). 

By  (d)  the  derived  function  is 

(x  -3){x^6)  +  (x  +  2){x-6)  +  {x  +  2)  {X  -  3), 
which  reduces  to  dx^  —  Ux, 

the  same  value  as  by  the  first  method. 

355.  Theorem  I.  When  the  derived  function  is  posi- 
tive, the  original  function  increases  with  x;  when  it  is 
negative,  the  function  decreases  as  x  increases. 

Proof,  When  we  increase  x  by  the  quantity  h,  Fx  is 
changed  to  F{x  -{-  1i),  and  is  increased  by  the  difference 

F{x  -\-h)  —  Fx. 
But,  by  {h)  and  (&'),  we  have 

F  {x  -\-  h)  —  Fx  =z  h  F'x  +  ¥  x  other  terms 

=z  h  {F'x  +  A  X  other  terms).       {e) 

Now  we  may  take  the  increment  h  so  small  that  h  x  other 
terms  shall  be  less  than  F'x,  and  then  F'x  -^  hx  other  terms 
will  have  the  same  sign  (+  or  — )  as  F'x. 

Then,  supposing  h  positive,  the  increment 

F{x  -\-  h)  —  Fx 

will  be  positive  when  F'x  is  positive,  and  negative  when  it  is 
negative. 

Theorem  II.  If  an  equation  has  equal  roots,  such  root 
will  also  he  a  root  of  the  derived  function. 


430  GENERAL    THEORY   OF  EQUATIONS. 

Proof.  Let  0  be  the  root  which  Fx  =:0  has  in  duplicate. 
Then  when  Fx  is  factored,  it  will  be  of  the  form 

Fx  =:  {x  —  a)  {x  -  P){x--P){x  —  y) {x  -  e). 

Now  when  we  form  F'x  by  method  (2),  the  factor  {x  —  |3) 
will  be  left  in  all  the  terms.  Therefore  x  —  fi  will  be  a  factor 
of  F'x,  Therefore,  when  x  =  13,  then  F'x  =  0,  so  that  P  is 
a  root  of  the  equation  F'x  =  0. 

356.  If  the  equation  Fx  =  0  contains  no  equal  roots,  and 
if  we  suppose  x  =  a  in  equation  (d),  all  the  terms  except  the 
first  will  vanish,  because  the  common  numerators  Fx  contain 
X  —  a  as  a  factor. 

In  the  case  of  the  first  term,  both  numerator  and  denomi- 
nator vanish  when  x  =  a;  therefore  we  must  find  the  limit  of 

when  X  approaches  «.     This  is  easy,  because 

Fx 
-— -  =  {x-(i){x-y) (z-  e). 

Therefore,  by  supposing  x  to  approach  a,  we  shall  have 

Fx 
Lim.  (x=a)  =  (a  —  p)  {a  ^  y)  ,  ,  .  ,  [a  —  e). 

Therefore,  by  changing  x  into  a  in  (d),  we  find 

F'a  =  {a-ti){a-y) (a  -  e). 

Hence 

The  derived  function  of  a  root  which  has  no  other 
root  equal  to  it  is  the  continued  product  of  its  difference 
from  all  the  other  roots. 

Significance  of  the  Derived  Function. 

357.  Theokem.  The  derived  function  expresses  the 
rate  of  increase  of  the  function  as  compared  with  that 
of  the  variable. 

Proof,    The  equation  {e)  may  be  expressed  in  the  form 
F(x  -f  li)  =  Fx-\-n  {F'x  +  Bh), 


FORM  OF  BOOTS,  431 

where  Bli^  is  the  sum  of  the  remaining  terms  of  the  develop- 
ment in  powers  of  h. 

We  then  have 

Increment  oix  =z  h. 

Corresponding  increment  of  Fx  =  F{x  -\-  h)  ^  Fx 

=:  h  {F'x  +  Bh). 

Eatio  of  these  increments,  — '^ -—■ =  F'x  +  Bh. 

If  we  suppose  the  increment  h  to  approach  zero  as  its 
limit,  the  product  Bh  will  also  approach  zero,  and  the  ratio  will 
approach  F'x  as  its  limit. 

But  this  ratio  of  the  increments  may  be  considered  as  the 
ratio  of  the  average  rate  of  increase  of  the  function  F  to  that 
of  the  variable  x. 

Hence,  when  we  plot  the  values  of  Fx  by  a  curve,  as  in 
§  345,  the  derived  function  shows  the  slope  of  the  curve  at 
each  point. 

When  the  derived  function  is  positive,  the  curve  is  running 
upward  in  the  positive  direction,  as  from  x=i—d  to  ^  =  0, 
and  from  x=i  -f5  to  x=z  +oo. 

When  the  derived  function  is  negative,  the  curve  slopes 
downward,  as  from  a;  =  0  to  x  =z  +4. 

When  the  derived  function  is  zero,  the  curve  at  the  corre- 
sponding point  runs  parallel  to  the  base  line,  as  at  0  and  +4f. 
If  this  point  corresponds  to  a  root  of  the  equation,  the  curve 
will  coincide  with  the  base  line  at  this  point,  and  will  there- 
fore be  tangent  to  it.     Hence,  from  §  356,  Th.  II, 

A  pair  of  equal  roots  of  an  equation  are  indicated  hy 
the  curve  touching  the  base  line  without  intersecting  it 

Forms  of  the  Roots  of  Equation. 

358.  Theorem  I.  Imaginary  roots  enter  an  equation 
with  real  coefficients  in  pairs. 

That  is,  \i  a  +  hi  be  a  root  of  such  an  equation,  then 
a  —  bi  will  also  be  a  root. 


432  GENERAL    THEORY  OF  EQUATIONS. 

Proof.     Let 

X^  -^p^X^-^  +  PzX^''^  + +  Pn-\X  -{-  pn  =  ^       (1) 

be  the  equation  with  real  coefficients,  and  let  us  suppose  that 
a  +  hi  is  a  root  of  this  equation.  If  we  substitute  a  +  hi  for 
X,  we  shall  have 

a;«  =  a«  +  na^-^  hi  —  !Ll!^lJ  ^/i-2  ^,2  _  ^  ^n-3  ^,3^-4.  etc. 

p^x^-^  =  p^a'^~^  +  p^  (n — l)a'*""%  —  etc. 
If  we  substitute  all  the  terms  thus  formed  in  equation  (1), 
and  collect  the  real  and  imaginary  terms  separately,  we  shall 
have  a  result 

A  +  Bi  =  0 

(§  324),  A  signifying  the  sum  of  all  the  real  terms, 
a^ ^^ ^  «^-2  yi^      p  ^  (^n-i^     etc., 

and  Bi  the  sum  of  all  the  imaginary  ones. 

In  order  that  this  equation  may  be  satisfied,  we  must  have 

identically 

A  =zO,     B  z=0    (§  327). 

Next  let  us  substitute  a  —  hi  for  x.  Since  the  even  powers 
of  hi  are  all  real,  and  the  odd  powers  all  imaginary,  this 
change  of  sign  will  leave  all  the  real  terms  in  (1)  unchanged, 
but  will  change  the  signs  of  all  the  imaginary  terms.  Hence 
the  result  of  the  substitution  will  be 

A  -  Bi. 

But  if  «  +  hi  is  a  root,  then,  as  already  shown,  ^  =  0 

and  ^  =  0 ;  whence 

A-  Bi=zO 

also,  and  therefore  a  —  hi  is  also  a  root. 

Def,  A  pair  of  imaginary  roots  v^hich  diflfer  only 
in  the  sign  of  the  coefficients  of  the  imaginary  unit  are 
called  a  pair  of  Conjugate  Imaginary  Roots. 

Theorem  II.  In  the  expression  Fx  every  pair  of  conju- 
gate imaginary  factors  form  a  real  product  of  the  second 
decree  in  x. 


DECOMPOSITION  OF  RATIONAL   FRACTIONS.        433 

Proof.     If  in  the  expression 

Fx  —  {x  —  a)  {x  —  (^){x  ^y)  .  .  ,  .  {x  —  s), 

we  suppose  a  and  i3  to  be  a  pair  of  conjugate  imaginary  roots, 
which  we  may  represent  in  the  form 

a  =1  a  -^  hi,        (3  =  a  —  hi, 
then  the  product  of  the  terms  {x  —  a)  {x  •—  h)  or  of 

{x  —  a  —  hi)  {x  —  a  +  hi), 
will  be  {x  —  of  4-  b\ 

or  x^  —  '^ax  -\-  a^  +  }^, 

a  real  expression  of  the  second  degree  in  x. 

Cor,  Since  Fx  can  always  be  separated  into  factors  of  the 
first  degree,  either  real  or  imaginary  (§  347,  Th.  I),  and  since 
all  the  imaginary  factors  enter  in  pairs  of  which  the  product 
is  real,  we  conclude : 

Every  entire  function  of  x  with  real  coefficients  may 
he  divided  into  real  factors  of  the  first  or  second  degree. 

Decomposition  of  Rational  Fractions. 

359.  Def.  A  Rational  Fraction  is  one  whicli  may 
Ibe  reduced  to  tlie  form 

ax>^  +  hxi^-^  +  cx!^-^  +  ....+/ 


X^  +  p^X^-^  +  p^X^-^  +  .  .  .  .   -{-  pn 

If  the  exponent  m  of  the  numerator  is  equal  to  or  greater 
than  the  exponent  n  of  the  denominator,  we  may  divide  the 
numerator  by  the  denominator,  obtaining  a  quotient,  and  a 
remainder  of  which  the  highest  exponent  will  not  exceed 
71-^1.    If  we  put 

fx,  the  numerator  of  the  above  fraction ; 
Fx,  its  denominator ; 
Q,  the  quotient; 
(px,  the  remainder ; 

"^x  d)X 

we  shall  have,    Rational  fraction  r=  4t-  —  (^  -f  -^*     f§  96.) 

Fx       ^  ^  Fx      ^^      ' 

28 


434  GENERAL    THEORY   OF  EQUATIONS, 

Q  will  be  an  entire  function  of  x,  with  which  we  need  not 
now  farther  concern  ourselves. 

The  problem  now  is,  if  possible,  to  reduce  the  fraction 

•^  to  the  sum  of  a  series  of  fractions  of  the  form 
Fx 

ABC  E 


x  —  a       X  ^  13       X  —  y       '  '  *  '  ~  x  —  e' 

A,  By  C,  etc.,  being  constants  to  be  determined,  and  a,  p,  y, 
etc.,  being  the  roots  of  the  equation  Fx  —  0.  Let  us  then 
suppose 

Fx'~x  —  a'^X'--li'^x  —  y'^ '^  x -•  e        ^^ 

Multiplying  both  sides  by  Fx,  we  have 

AFx        BFx         CFx  EFx 

X  —  a       X  —  P       x  —  y  x  —  E        ^   ' 

We  require  that  this  equation  shall  be  an  identical  one, 

true  for  all  values  of  x.     Let  us  then  suppose  a;  =  «.     Then 

because  by  hypothesis  cc  is  a  root  of  the  equation  Fx  =  0,  we 

have  Fa  =  0,  and  the  terms  in  the  second  member  will  all 

vanish  except  the  first.     If  there  is  only  one  root  «,  we  have 

(§  357), 

^.         Fx  ^, 

Lim.  {x=a)  =  Fa. 

X  —  a 

Therefore,  changing  x  to  a,  we  have 

(pa  =  AF'a, 

which  gives  A  =  ~j — 

In  the  same  way  we  may  find 

B--^  (c) 

etc.      etc. 

Sttbstittttiiig  these  values  of  A,  B,  etc.,  in  tlie  equation  {b), 
it  becomes 


DECOMPOSITION   OF  RATIONAL    FRACTIONS.        435 

_0X  _  (Pa (t>?     _  (t>y  , 

i^o;  ~  \x  -  a)  Fa  ^  {x  ~  fi)  F'[3  '^  {x  -  y)  F'y  "^     '^• 

Note.  The  critical  student  should  remark  that  in  the 
preceding  analysis  we  have  not  proved  that  the  expression  of 
the  rational  fraction  in  the  form  ih)  is  always  possible,  but 
have  only  proved  that  if\i  be  possible,  tlien  the  coefficients  A^ 
B,  G  must  have  the  values  (c).  To  prove  that  the  form  is 
possible,  the  second  member  of  Q))  may  be  reduced  to  a  com- 
mon denominator,  which  common  denominator  will  be  Fx, 
and  the  sum  of  the  numerators  equated  to  (px.  By  equating 
the  coefficients  of  the  separate  powers  of  x,  we  shall  have  n 
equations  to  determine  the  n  unknown  quantities  A,  B,  G, 
etc.  Since  n  quantities  can,  in  general,  be  made  to  satisfy  n 
equations,  values  of  J,  B,  G,  etc.,  will  in  general  be  possible. 

It  will  be  instructive  to  solve  the  following  exercises,  both 
directly  and  by  the  comm.on  denominator. 


I.  Decompose 


EXAMPLES. 

2a;2  _  3a;  +  5 


a^^^x^j^  36 


We  have  already  found  the  roots  of  the  denominator  to  be 
-  2,  3,  and  6.    Using  the  formulas  (c),  we  find 

(t>x  =  2x^  —  dx-\-5, 

Fx  =  x^—  7x^  H-  36  =  {x-{-  2)  {x  —  3){x^  6), 
F'x  =  Sx^  —  Ux; 

(pa  =  19,  (pp  =  14,  0y  =  59 ; 

F'a  =  40,  F'P  =  -  15,      F'y  =  24. 

2x^  —  3x  +  5 19^ 14  59 

x^  —  W  +  36  ""  40  (x  +  2)  '~  \h{x  —  3)  "^  24  (a;  —  6)' 

_  2^3  —  7^  +  3  2x^  —  lx-^3 

2.  Decompose  —  ' 


x^  —  2x^  —  X  -^2        (^+1)  (^—1)  {x—2) 

Here  the  roots  of  the  denominator  are  —  1,  1,  and  2.     Let 
us  effect  the  decomposition  by  the  following  method.    Assume 


436  GENERAL    THEORY  OF  EQUATIONS, 

2:^2-7^  +  3  A       ^       B  G 


{x  +  l){x  —  \){x-'2)       x-\-l^  x  —  1^  x  —  ^ 

Reducing  the  second  member  to  a  common  denominator, 
it  becomes 

A  {x^  —  3a;  +  2)  +  ^  (^^  —  ^  —  2)  +  0  {x^  —  1) 
{x  J^l)(x  —  l){x  —  ^)~ 

Since  both  members  now  have  the  same  denominator,  their 
numerators  must  also  be  equal.  Equating  them,  after  arrange 
ing  the  last  one  according  to  powers  of  x,  we  have 

{A^B-^C)x^-  {dA-\-B)x-^^A—'^B--C—  2.7:2.-7^  +  3. 

Since  this  must  be  true  for  all  values  of  x,  we  equate  the 
coeflBcients  of  a;  in  each  member,  giving 

A  -\-  B  -^  C  =  ^, 

'dA-\-  B  =  1, 

2A-2B-  C=  3. 

These  equations  being  solved  give 

A  =  2,        B  z=l,        (7  =:  —  1. 

Substituting  in  (d), 

2a;2  —  7a:  4-  3  2  1  1 


(a:  +  1)  (a;  —  1)  (a;  —  2)        a;  +  1  "^  a:  —  1       x  —  2 

EXERCISES. 

Decompose : 
a^jMO 

2a^  —  12a^  —  8ar  +  12 
^*  x^-^ox^^fl 

2a 


x^  4-  Sx  -\-4: 

x^  -{.  x^  —  4.x  —  4: 

X 

x^-^a^ 

aW 

(a;2  _  «2)  {^  -  h^) 

x^  —  w^ 

360.  When  the  equation  Fx  =zO  has  two  or  more  equal 
roots,  the  preceding  form  fails,  because  all  the  terms  of  the 
second  member  of  {h')  will  then  vanish  when  we  suppose  x 
equal  to  one  of  the  multiple  roats.  In  this  case  we  must  pro- 
ceed as  follows : 


DECOMPOSITION   OF  RATIONAL    FRACTIONS.        437 

If  Fx  =  {x  —  a)^  {x  —  0)^  (x  —  y)P, 

we  suppose 

iric        (:^  _  a)^  '^  {x—  cc)^-^  '^  (x  —  «)'^-2  "^  "^  o;  —  €« 

+  (^  _  /3)^  +  J^'^yi^i  + +  ^z:^ 

+ ^ + ^-1 4-  4.  _^^. 

^  {x  —  y)P  ^  {x-^  y)P-i  ^     *  *  '  ^  ^  —  y 

etc.  etc.  etc. 

In  the  case  of  m,  n,  or  p  =  1,  this  form  will  be  the  same 
as  (b),  as  it  should. 

By  reducing  the  second  member  to  a  common  denominator, 
and  equating  the  sum  of  the  numerators  to  (px,  we  shall  have, 
as  before,  a  number  of  equations  the  same  as  the  degree  of  x 
in  Fx. 

EXAMPLE. 

^  Sx^  —  9x^  —  2:r  —  1 

Decompose 


x^  _  2x^  —  2x^  -^  4:X^  -^  X  —  2' 

of  which  the  roots  of  the  denominator  are  —  1,  —  1,  1,  1,  2. 

Solution.    Because  of  the  roots  just  given,  the  expression 
to  which  the  fraction  is  to  be  equal  is 

A  A,  B  B.  G 


(x^lf^x--l^(x-{-lf^x-\-l^x  —  2 

Eeducing  to  a  common  denominator,  and  equating  the  co- 
efficients of  the  powers  of  x  to  the  coefficients  of  the  corre- 
sponding powers  in  the  numerator  8::^^  —  4cX^  —  2a:  —  1,  we 
have 

A^-{-B^-{-C=      0, 

—  Ji  +  ^  -3^1  +  ^  =       8, 
—  3J[i  +  ^1  -  4^  —  2C  =  —  9, 

2A^  —2A^2B^-\-2B+C=—l. 
Solving  these  equations,  we  find, 

A    =       1,  B    =       2,  0=3. 

A,  =-2,  B,=.-l, 


438  GENERAL    THEORY  OF  EQUATIONS. 

The  given  fraction  is  therefore  equal  to 


{x  —  iy      x  —  l'{x-\-lY      x-j-1    '  x^  2 


EXERCISES. 


Decompose  ^^^-         ^ns.  ^  +  j;J-jy,. 

x-1  x^-2 

3- 


•        {X  +  1)2  '^'       x^  —  X^-X  +  1 

x-^2 

'^'     cx^  -\-~x^  —  x  —  1 

Greatest  Common  Divisor  of  Two  Functions. 

361.  When  we  have  two  equations,  some  values  of  the 
unknown  quantity  may  satisfy  them  both.  They  are  then  said 
to  have  one  or  more  common  roots.  Such  equations,  when 
factored  as  in  §  347,  will  have  a  common  factor  or  divisor  for 
each  common  root.     Hence, 

Theorem.  The  common  roots  of  two  equations  majj 
be  found  frovv  their  greatest  comjmon  divisor. 

Problem.  To  find  the  greatest  cowymon  divisor  of  two 
equations. 

This  problem  is  solved  by  dividing  the  two  polynomials  by 
the  methods  of  §§  96,  97,  and  232. 

Example  i.    To  find  the  greatest  common  divisor  of  the 
two  polynomials, 

and  x^—2x^  +  4:X^  +  2:^;  —  5. 

FIRST  DIVISION. 

a;5  _  4^  ^  12a;3  ^  4^2  _  13^  I  x^  —  2a^  +  ^x^  +  2x  —  5 

a;5  _■  2x^  +    4a^  +  2x^  --    6x  \  x  —2 

—  2a;4+    8.^3+  2x^  —    Sx 

—  2x^  4-    4:X^  —  Sx^  —    4a;  +  10 


4:X^  +  10^:^  —    4:X  —  10  =  first  remainder. 


GREATEST  COMMON  DIVISOR. 


439 


SECOND  DIVISION. 

-  4t 

-10 

x^  +  I- ^^  —      i?:2  _  1^ 

i^  -1 

-^x^+    6x^-{-%x-6 

^:^;2  «_^  —  second  remainder; 

or,  5^(:r2  —  1)  =  second  remainder. 

In  the  next  division,  we  may  omit  the  fractional  factor  ^-, 
because  every  value  of  x  which  satisfies  the  equation  x^ — 1  =z  0 
will  also  make  -^  {x^  —  1)  ~  0,  so  that  these  two  equations 
have  the  same  roots.  In  this  process  we  may  always  multiply 
or  divide  the  terms  of  each  remainder  by  any  factor  which  will 
make  their  coefficients  entire. 


THIKD  DIVISION. 

4:X^  +  10:^2  —  4:X  —  10 

x^-1 

4,0?              ^4,x 

4.x  +10 

10a;2            — 10 
10a;2            — 10 

0 


0 


Hence,  the  G.C.D.   of  the   two  functions  is  x^  —  1,  and 
their  common  roots  are  +1  and  —1. 

This  result  may  also  be  reached  by  factoring  the  given 
equations,  and  multiplying  the  common  factors,  thus : 

a^  -_  4^4  _|_  i22;3  +  4a;2  —  13a; 

=  X  {x  —  1)  {x  +  1)  {x  -'  2  —  di)  {x  —  2  +  30, 

x^^2x^-\-  4a;2  +  2:^;  —  5 

=  {x  —  l){x  +  l){x^l--  2i)  (a;  -  1  +  2i), 

We  see  that  the  common  factors  are 

(x  -~  1)  (r^;  +  1)  =  a;2  -  1. 


440  GENERAL    THEORY   OF  EQUATIONS. 

The  rules  for  throwing  out  factors  from  divisor  or  dividend 
are  as  follows : 

I.  If  both  given  polynoTnials  contain  the  same  factor 
in  all  their  terms,  re7)%ove  this  factor,  and  after  the 
G.  C.  D.  of  the  remaining  factors  of  the  two  polynomials 
is  found,  multiply  it  by  this  factor. 

Proof.  If  a  be  such  a  factor^  and  X  and  Y  the  quotients 
after  this  factor  is  removed  from  the  two  polynomials,  the  lat- 
ter, as  given,  will  be 

aX    and    a  Y. 

Since  a  is  now  a  common  divisor  of  both  given  polynomials, 
if  we  call  D  the  G.O.D.  of  Xand  Z,  it  is  evident  that  aD  will 
be  the  G.C.D.  of  aX  and  aY. 

II.  Any  factor  common  to  all  the  terms  of  any  divi- 
sor, and  not  contained  in  the  dividend,  ma;y  be  thrown 
out. 

Proof.  If  this  factor  were  any  part  of  the  G.C.D.  sought, 
it  would,  by  §  232,  be  a  factor  of  each  dividend.  Since  the 
only  factors  we  require  are  those  of  the  G.C.D,  factors  In  a 
divisor  only  may  be  rejected. 

EXERCISES. 

Find  the  G.C.D.  of  the  following  polynomials: 
I.     x^  —  1  and  ^  —  1. 

2 

3 


x^  —  1  and  x^  —  1. 

«5  _  2a^  —  ^3  +  3«3  _  2a— 15  and  a^^a^—^a^—a  +  6. 

26x^  -j-6x^  —  x  —  l  and  20a;4  -\-x^  —  l. 

a^  +2^2  +  9  and  a^  +  2a^  _  6a  —  9. 

m^  -\-  3m^  +  3m  +  1  and  m^  —  1. 

xi^Sa^  +  21.^2  _  20:r  +  4  and  2c(^  —  12a;2+21i2^~10. 

a^  -{-  a^  —  a  —  1  and  a^  +  a^  —  a  —  1. 


363.  The  given  polynomials  may  be  functions  of  two 
or  more  symbols,  as  in  §  97.  We  then  arrange  them  accord- 
ing to  the  powers  of  one  of  the  symbols,  and  perform  the  divi- 
sions by  the  precepts  of  §  97.  I 


GREATEST  COMMON  DIVISOR.  441 

Ex.     Find  the  greatest  common  divisor  of 

q:^  —-  ax^  +  a  {b  -{-  c)  X  —  ahc  —  hx^  —  cx^  +  hex 
and        01?  —  ax^  —  a{b  +  c)x  —  abc  +  hx^  -\-  cx^  +  hex. 

The  quotient  of  the  first  division  will  be  unity,  so  we  write 
the  two  functions  under  each  other,  thus : 

a?  —       {a  +  h  -{-  c)x^  -^  {ah  -\-  he  -^  ca)x  —  ahc 
a?  +  {—  a  -^h  -^  e)x^  —  {ah  —  he  -\-  ca)  x  —  ahe 

—  2{h  +  e)x^+        2  (a^  +  ae)  x  =  1st  rem. 

Dividing  this  remainder  by  —2{h+  e),  we  have  the  next 
divisor.     We  then  perform  the  next  division  as  follows : 


a?  +  {—a-^h  +  e)x^  —  {ah—he+ea)  x  —  ahe 
x^  —  ax^ 


x^  —  ax 


X  -V^b^rc) 


{b-\-G)  x^  —  {ah—hC'\-ca)  x  —  ahe 
{h  -\-e)x^  —  {ah         +  ca)  x 


hex  —  ahe  =  2d  rem. 

Dividing  this  by  the  factor  he,  which  is  contained  in  all  its 
terms,  we  have  x  —  a  for  the  next  divisor,  which  we  find  to 
divide  the  last  divisor,  and  therefore  to  be  the  G.C.D.  required. 

'EXERCISES. 

Find  the  G.C.D.  of 


a^  +  Shcx-h  ¥  —  €^  and  a?  4-{c-^h)  x^-i-{P'i-he+(?)  x, 

a?  +  Sax  -\-  a^  —  1  and  x^  —  {a^  —  2a)  x  -{-  a  —  1. 

{a-{-h-{-  c)  {ah  +  ^c  +  ca)  —  ahe  and  a^  -j-  ah  —  ae  —  he, 

x^  +  4a^  and  x^  —  2a^x  -f-  4:aK 

x^  —  ax^  —  ¥x  +  ah^  and  x^  —  a^, 

x^  -\-  a^  -^  ¥  —  dahx  and  x^  +  2ax  +  a^  —  5^. 

^4  _  2a;2  +  2  -  -,  +  ^  and  0:4  __  2^:2  +  -o  -  1^. 

X^         X^  0^         7^ 

x^  —  x^y  +  xy^  —  y^  and  x^  -\-  x^y^  +  y^. 


442  GENERAL    THEORY  OF  EQUATIONS. 

Traiisformation  of  Equations. 

363.  Def.  An  equation  is  said  to  Tbe  Transformed 
when  a  second  equation  is  found  whose  roots  bear  a 
known  relation  to  those  of  the  given  equation. 

Rem.  Sometimes  we  may  be  able  to  find  a  root  of  the 
transformed  equation,  and  thence  the  corresponding  root  of 
the  original  equation,  more  easily  than  by  a  direct  solution. 

Problem  I.  To  change  the  signs  of  all  the  roots  of  an 
equation. 

Solution,  By  changing  x  into  —  ic  in  a  given  equation, 
the  signs  of  the  terms  containing  odd  powers  of  x  will  be 
changed,  while  those  of  the  even  powers  will  be  unchanged. 
Hence,  if  a  be  any  root  of  the  original  equation,  —  a  will  be 
a  root  of  the  equation  after  the  signs  of  the  alternate  terms  are 
changed.     Hence  the  rule : 

Change  the  signs  of  the  alternate  terms,  of  odd  and 
even  degree,  in  the  equation. 

Problem  II.  To  diminish  all  the  roots  of  an  equa- 
tion hy  the  same  quantity  h. 

Solution,     If  the  giyen  equation  is    . 

x^  +  p^x^-^  +  p^xP'-^  + -\-pn  =  0, 

and  if  y  is  the  unknown  quantity  of  the  required  equation,  we 
must  have 

y  =.  X  --  h. 

Therefore,  x  :=:  y  -{-  h. 

Substituting  this  value  of  x  in  the  equation,  it  will  become 


?/''-i-(i?i+^/02/''"H 


^,^  +  (^_l)^^7,  +  g)/,2  l^^-2  +  etc.   {a) 


When  h,  n,  and  the  ^'s  are  all  given  quantities,  the  coeffi- 
cients of  y  become  known  quantities. 


I 


GENERAL    THEOET   OF  EQUATIONS.  443 


EXERCISES. 

1.  Transform  the  equation  x^  —  ^x  —  ^  z=z  0  into  one  in 
which  the  roots  shall  be  less  by  1. 

2.  Transform  x^  —  dx^  -\-  blx  —  1  =  0  into  one  in  which 
the  roots  shall  be  greater  by  5. 

364:,  Removing  Terms  from,  Equations,  The  quantity  li 
may  be  so  chosen  that  any  required  term  after  the  first  in  the 
transformed  equation  shall  vanish.  For,  if  we  wish  the  second 
term  of  the  equation  {a)  to  vanish,  we  have  to  suppose 

j)^  +  nil  =z  0, 
which  gives  ^  =  —  ^. 

We  then  substitute  this  value  of  Ji  in  the  equation  {a), 
which  gives  an  equation  in  which  the  second  term  is  wanting. 

If  we  wish  the  third  term  to  vanish,  we  must  determine  h 
by  the  condition 

which  requires  the  solution  of  a  quadratic  equation.  Each 
consecutive  term  is  one  degree  higher  in  the  unknown  quan- 
tity h,  and  the  last  term  is  of  the  same  degree  as  the  original 
equation. 

This  method  is  principally  applied  to  make  the  second 
term  disappear,  which  requires  that  we  put 

n 

Example.  Make  the  second  term  disappear  from  the  fol- 
lowing equation, 

x^  ■\-  px  -^  q  •=:  0, 

Solution,     Hence,  w  =  2  and  p^  =  p,  so  that 

;.  =  -|. 

P 


444  GENERAL    THEORY    OF  EQUATIONS. 


f-P    +g  =  0, 


Making  this  substitution,  the  equation  becomes 

4 

which  i«  the  required  equation. 

Eem.  This  process  affords  an  additional  elegant  method  of 
solving  the  quadratic  equation. 
The  last  equation  gives 


/p2  1 

The  value  of  x,  being  equal  to  i/  +  h,  then  becomes 
which  is  the  correct  solution. 

EXERCISES. 

Eemove  the  second  term  from  the  following  equations  : 

1.  x^—%X^-\-^X—l=:  0. 

2.  ii;4  _  4^3  _|_   3^2  _  8    zz:    0. 

3.  ^  -^  5^4  +  2a;3  _!_  2^2  _  3^  __  0, 

4.  x^  —  12^5  4-  2:^2  —  X  =z  0. 

Eem.  The  theory  of  the  above  process  will  be  readily  com- 
prehended by  recalling  that  the  coefficients  of  the  second  term 
is  equal  to  the  sum  of  the  roots  taken  negatively,  or  if  a,  P,  y, 
etc.,  be  the  roots, 

It  is  evident  that  if  we  subtract  the  arithmetical  mean  of 

all  the  roots,  that  is,  —  — ,  from  each  of  them,  their  sum  will 
vanish,  because 

ec4-^  +  /J+^  +  y +  ^  +  etc.  =  ~i?i  +n^  =  0. 
n  n        '        n  ^^  n 

Hence,  when  we  put  «/  —  —  for  ic  in  the  equation,  the  sum 
of  the  roots,  and  therefore  the  second  term,  vanish. 


GENERAL    THEORY   OF  EQUATIONS.  445 

365.  Problem.  To  transform  an  equation  so  that  the 
roots  shall  he  mivltiplied  by  a  given  factor  m. 

Solution,  Since  the  roots  are  to  be  multiplied  by  m,  the 
new  unknown  quantity  must  be  equal  to  mx.  So  if  we  call 
this  quantity  y,  we  have 

y  z=  mXy 

which  2jives  x  ^=l  —* 

Substituting  this  in  the  general  equation,  it  becomes 

^  +  Pi  —^1  +  P2  —^2  +  •  •  •  •   +2Jn  =  0, 

Multiplying  all  the  terms  by  m%  the  equation  becomes 

yn  _^  7np^y^~^  +  m^p^y^-^  +  ....  +  yn'^pn  =  0. 
Hence  the  rule,^ 

Multiply  the  coefficient  of  the  second  term  by  m,  that 
of  the  thii^d  by  m^,  and  so  on  to  the  last  term,  which  will 
be  multiplied  by  m^. 

If  the  roots  are  to  be  divided,  we  divide  the  terms  in  the 
same  order. 

EXERCISES. 

1.  Make  the  roots  of  ri;^  —  2a;  +  3  =  0  four  times  as  great. 

2.  Divide  the  same  roots  by  2. 

366.  Problem.  To  transform  an  equation  so  that  its 
roots  shall  be  squared. 

Solution.    Let  the  given  equation  be 

a;4  +  PxX^  +  p^x'^  +  l^x  +  p^  =:  0. 

If  ^  be  the  unknown  quantity  of  the  new  equation,  we 
must  have 

y  —  x\ 

which  gives  x  =:  ±2/^. 

If  we  substitute  a;  =  ^/^  in  the  given  equation,  It  may  be 
reduced  to  the  form 

y'^  4-  P%y  +  ^4  +  {PiV  +  Pz)  r  -  0. 


446  GENERAL    THEORY   OF  EQUATIONS. 

If  we  substitute  a;  =  —  y^,  the  result  will  be 

y'^  +  ihv  -^-v^  —  {Piy  +  p-i)  r  =  0- 

Since  the  value  of  y  must  satisfy  one  or  the  other  of  these 
equations,  it  must  reduce  their  product  to  zero  ;  we  therefore 
multiply  them  together.  Considering  them  as  the  sum  and 
difference  of  a  pair  of  expressions,  the  product  will  be 

{y^  -^p^y  +  V4I  -  {Piy  ^  p^Yy  =  0, 

or 

y'+{^P2-Pi')y'+{P2'+^P4-^PiP3)f+(^:P2P^'-Pz')y+p,' 

=  0. 

EXERCISES. 

1.  Transform  the  quadratic, 

x^  —  5x  -\-  6, 

of  which  the  roots  are  2  and  3,  into  an  equation  in  which  the 
roots  shall  be  the  squares  of  2  and  3,  using  the  above  process. 

2.  Transform  in  tlie  same  way 

^3  ^  12:^2  _!_  44^.  _!_  43  _  0, 

3.  Transform 

x^  —  4:X^  —  10ir3  +  40a;2  -{.  dx  —  36  =  0. 

Generalization  of  the  Preceding  Problems. 

367.  Problem.  Given,  an  equation  of  any  degree 
in  an  unhnoivn  quantity  x ; 

Required,  to  transform  this  equation  into  another  of 
which  the  root  shall  he  a  given  function  of  x, 

Solution.  Let  ?/  be  a  root  of  the  required  equation,  and  fx 
the  given  function.     We  must  then  have 

f^  =  y' 

Solve  this  equation  so  as  to  obtain  x  as  a  function  of  y. 
Substitute  this  value  of  x  in  the  original  equation,  and  form  as 
many  equations  as  there  are  values  of  y. 

The  product  of  these  equations  will  be  the  required  equa- 
tion in  «/. 


GENERAL    TUEOBT   OF  EQUATIONS.  447 

EXERCISES. 

1.  Transform 

iC2  _  7x   +   10   =:   0 

SO  that  the  roots  of  the  new  equation  shall  be  '^x\ 

2.  Transform        x^  —  ^x^  +  2cc  =  0 
so  that  the  roots  shall  be  ax  +  h. 

3.  Transform         x^  —  ^x  -\-  1^  =:  0 

so  that  the  roots  shall  be  ^x^'  —  3. 

o 

Resolution  of  Nvimerical  Equations. 

368.  Convenient  method  of  computing  the  numerical  value 
of  an  entire  function  of  x  for  an  assumed  value  of  x. 

If  we  have  the  entire  function  of  x, 

Fx  =  ax^  +  bx^  +  cxP  -^  dx  +  e, 
we  may  put  it  in  the  form 

Fx  =1  "^[^(^ax  -\-  l))x  -{■  c'\x  +  d\  X  -\-  e. 
Therefore^  if  we  put 

ax  -\-h  ^  V ,  h'x  -\-  c  =:  c\ 

c'x  +  c?  =:  d',  d'x  -{-  e  =^  e\ 

we  shall  liave  Fx  =  e\ 

Numerical  Example.     Compute  the  values  of 
Fx  =  2qP  —  ^x!^  —  6a;3  +  82)  —  9 

f or  a;  =:  3  and  a;  =  —  2. 
We  arrange  the  work  thus : 


Coefficients, 
Prod,  by  {x=^), 

2 

-3         -6 

+  6         +9 

~+3         +3 

0 

+  9 

+  9 

+   8 

+  27 
+  35 

-     9 

+  105 
+   96 

Hence, 

F^  =  96. 

For  2:  =  —  2, 

2 

-3       -  6 
—4       +14 

-7       +8 

0 
—16 
—16 

+  8 

+  32 

+  40 

—  9 
—80 
-89 

Hence, 

F(~2)^  - 

89. 

448  GENERAL    THEORY   OF   EQUATIONS, 

This,  it  will  be  noticed,  is  a  more  convenient  process  than  that  of 
forming  the  powers  of  x  and  multiplying  and  adding. 

369.  Having  an  entire  function  of  x,  and  jMtting  x:=.r-\-li^ 
it  IS  required  to  develop  the  function  in  poiuers  of  lu 

It  will  be  remarked  ihat  this  problem  is  substantially  identical  with 
that  of  §  362,  and  the  solution  of  this  will  be  the  solution  of  the  formero 
But  in  the  former  case  h  was  supposed  to  be  a  given  quantity,  whereas  it 
is  now  the  unknown  quantity  corresponding  to  y  in  the  former  problem. 

Example  of  the  Problem.     If  we  have  the  expression 

Fx  —  2x^  +  3a;2  +  4, 

and  put  X  =  2  -\-  h,  it  will  become,  by  developing  the  sepa- 
rate terms, 

F  {2  -\-h)  =  2¥  +  loh^  +  367^  +  32. 

Ge:n"eral  Rule  for  the  Process.  First  compute  the 
value  of  Fr  by  the  process  employed  in  §  366. 

Then  repeat  the  process,  using  the  successive  sums  ob- 
tained in  the  first  process  instead  of  the  corresponding 
coefficients,  and  stopping  one  term  before  the  last.  The 
result  will  be  the  coefficient  of  li. 

Repeat  the  process  with  the  new  sums,  stopping  yet 
one  terin  sooner.     The  result  luill  be  the  coefficient  of  W. 

Continue  the  repetition  until  ive  have  the  first  term 
only  to  operate  upon,  which  will  itself  be  the  coefficient 
of  the  highest  power  of  h, 

Ex.  I.     The  example  above  given  is  performed  as  follows: 


Coefficients, 

+  2 

+  3 

0 

+  4 

Product  by  r, 

4 

14 

28 

First  sums, 

7 

14 

32 

Second  products, 

4 

22 

Second  sums, 

11 

36 

Tliird  product, 

4 
15 

Result,         F{2  +  h) 

1  ^  2h^+inm+mh+ 

32. 

Ex.  2.     In  the  function^ 

Fx  =  2x^  —7x^  +  52^  —  2x^  ^  Qx  —  8, 
let  us  put  X  =1  3  -{-  Ji,  and  express  the  result  in  powers  of  y^. 


GENERAL    THEORY   OF  EQUATIONS.  449 


CoefEcients,             2 
Products  by  3, 
First  sums. 
Second  products, 

-7 

6 

-1 

+  6 

+  5 
-3 

+  2 
+  15 

-2 

+  6 

+  4 

+  51 

+  6 

+  12 

+  18 
+  165 

-8 

+  54 

+  46 

Second  sums, 
Third  products, 

+  5 
6 

+  17 
33 

+  55 
150 

183 

Third  sums. 

11 
6 

17 
6 

23 

50 

51 

101 

205 

Result,        F{^-]-h)  =  2h'  +  237^^  +  lOl/i^  +  205^^  + 183^  +  46. 


EXERCISES. 


1.  Compute  2A5  _^  23/^4  _|_  101/^3 _^  205/^2 -f  183^ +  46,  when 
h  =  x  —  d. 

2.  Compute  a^  —  "^/x  +  7  f or  i^;  =  —  4  +  7^,  —  3  +  k,  etc., 
to   +3  4-  h. 

Proof  of  the  Preceding  Process.  If  we  develop  the  ex- 
pression 

a{h  +  r)''  -\-'b{n-\-rY-^  +  c(7i  +  r)^-2  _^  d{7i-^rY-^  +  etc., 
and  collect  the  coefficients  of  like  powers  of  h,  we  shall  find 

Coef.  of  7^^     —  a, 

li^-^  r=r  nar  +  h, 

nn-'i  =  {^\  ar'^  H-  {n  -\)lr  -\-  c,  {A) 

in-^  =  [|]  ar^  +  (--^)  br^  +  (^^  -  ^)  cr  +  d, 

Now  examining  Ex.  2  preceding,  it  will  be  seen  that  we  can 
make  the  computation  by  columns,  first  computing  the  whole 
left-hand  column  and  thus  obtaining  the  coefficient  of  h^~^, 
then  computing  the  next  column,  thus  obtaining  the  coeffi- 
cient of  h^~%  and  so  on.  Commencing  in  this  way,  and  using 
the  literal  coefficients,  a,  b,  c,  etc.,  and  the  literal  factor  r,  we 
shall  have  the  results : 


450  GENERAL    THEORY   OF  EQUATIONS. 

ah  c                                                  j 

ar  ar^  +  br 

ar  -\-  b  ar^  -{-    hr  -{-  c 

ar  2ar^  +  br 

2ar  +  b  dar^  +  2br  +  c 

ar  3ar^  +  br 

3ar  +  b  6ar^  +  3br  +  c 

nar  -^  b  (9)  ^^^  +  {n  —l)br-[-  c. 

If  n  is  the  degree  of  the  equation,  then,  by  the  preceding 
process,  we  shall  add  the  product  ar  to  b  n  times,  the  n  sepa- 
rate sums  being 

ar-\-b,     2ar-\-b,     3ar-\-b,  ....  nar-{-b. 

To  form  the  second  column,  we  multiply  each  of  these 
sums  except  the  last  by  r,  and  add  them  to  the  coefficient  c. 
The  terms  in  ar  added  being  ar^,  2ar^,  3ar%  etc.,  the  sum 
will  be  (1  +  2+3  + -\-7i  —  l)  ar^.     The  coefficient  is  a  figu- 

ni   (^  __  '^\ 

rate  number  equal  to  — ^^-^r (§§  286,  287).     The  sum  of 

2 

the  coefficients  of  br  is  n  —  1,  because   there   are  n — 1  of 

them  used,  each  equal  to  unity.     Therefore  the  final  result  is 

( -  j  ar'^'  +  {n  —  l)br  -\-  c,  - 

which  we  have  found  to  be  the  coefficient  of  /^^~^. 

In  this  second  column  the  partial  sums  or  coefficients  of 

ar'^  are  ^ 

1,    1  +  2  =  3,    1  +  2  +  3=:  6,  etc.,  to    1  +  2  +  3  +  .... +(/^— 2)= 

Therefore  the  numbers  successively  added  to  form  the  co- 
efficients of  ar^  in  the  third  column  are  1,  1  +  3  =  4,  1  +  3  +  6 
=  10,  etc.  The  coefficients  of  br'^  will  be  the  same  as  those  of 
ai^  in  the  column  next  preceding. 

Continuing  the  process,  we  see  that  the  coefficients  uro 
formed  by  successive  addition,  as  in  the  following  table,  whero 
each  number  is  the  sum  of  the  one  above  it  plus  the  one  on  its 


GENERAL    THEOBT   OF  EQUATIONS.  451 

^  y.  |.2  ^  y.4  ^  y.8        q\^q^ 


w 

1 

1 

1 

1 

1 

1     etc. 

h 

2 

3 

4 

5 

6 

etc. 

A2 

3 

6 

10 

15 

etc. 

h^ 

4 

10 

20 

etc. 

U 

5 

15 

etc. 

¥ 

6 

etc. 

h^ 

etc. 

etc. 

etc. 

left.  We  have  carried  the  table  as  far  as  nz=z%  and  the  ex- 
pressions at  the  bottom  of  each  column  will,  when  7^  =  6,  be 
formed  from  the  numbers  in  this  table,  taken  in  reverse  order, 
thus : 

Column  under  hy    Qar  +     b ; 

^'  "     c,  15ar^+  hh*  -f-  c; 

«  ''     d,  20ar^ + lObr^  -{-4:cr  +  d; 

"  ''     e,  15ar^  +  10br^-\-dcr^+3dr  +  e; 

''  ''     f,     6ar5+  5Z»r4  +  4cr3  4-3^r2  +  26r+/; 

■"  "     g,      ar^-j-     br^-i-  cr^-\-  dr^+  er^ +//•+?/. 

IS^ow  the  numbers  of  the  above  scheme  are  the  figurate 
numbers  treated  in  §  287,  where  it  is  shown  that  the  n^^  num- 
ber in  the  i^  column  after  the  column  of  units  is 

n(n  +  1)  (^^  +  2) (n  -{-  i  —  1)  _  /nj-i-^n 

1.2-3 i  "~  \         i        /' 

Comparing  with  the  coefficients  in  the  equations  (A),  we 
see  that  the  two  are  identical,  which  proves  the  correctness  of 
the  method. 

Sao.  Application  of  the  Preceding  Operation  to  the  Ex- 
traction of  the  Roots  of  Numerical  Equations*  Let  the  equa- 
tion whose  root  is  to  be  found  be 

a^  +  Ix^-'^  +  cx'^-^  4-  .  .  .  .  -^  g  —  0. 

We  find,  by  trial  or  otherwise,  the  greatest  whole  number 
in  the  root  x.     Let  r  be  this  number.     We  substitute  r-]-h  for 


452  GENERAL    THEOllT   OF  EQUATIONS. 

X  in  the  above  expression,  and,  by  the  preceding  process,  get 
an  equation  in  /?,  which  we  may  put  in  the  form 

ali^  -f  Vli^-^  +  cVi^-^  +  d!li'^-^  + 4-  ^'  =  0. 

Let  /be  the  first  decimal  of  li.  We  put  r'  +  /^'  for  7^  in 
this  equation,  and,  by  repeating  the  process,  get  an  equation 
to  determine  h\  which  will  be  less  than  0.1.  If  r"  be  the 
greatest  number  of  hundredths  in  li\  we  put  li!  =  r" -\-h",  and 
thus  get  an  equation  for  the  thousandths,  etc. 

371.  The  first  operation  is  to  find  the  number  and  approx- 
imate values  of  the  real  roots.  There  are  several  ways  of  doing 
this,  among  which  Sturm's  Theorem  is  the  most  celebrated, 
but  all  are  so  laborious  in  application  that  in  ordinary  cases  it 
will  be  found  easiest  to  proceed  by  trial,  substituting  all  entire 
•  numbers  for  x  in  the  equation,  until  we  find  two  consecutive 
numbers  between  which  one  or  more  roots  must  lie,  and  in 
difficult  cases  plotting  the  results  by  §  345. 

It  is,  however,  necessary  to  be  able  to  set  some  limits  be- 
tween which  the  roots  must  be  found,  and  this  may  be  done 
by  the  following  rules : 

I.  An  equation  in  which  all  the  coefficients,  including 
the  ahsoliite  term,  are  positive,  can  have  no  positive  real 
root. 

For  no  sum  of  positive  quantities  can  be  zero. 

II.  If  in  computing  the  value  of  Fx  for  any  assumed 
positive  value  of  x,  hy  the  process  o/  §  366,  we  find  all  the 
sums  positive,  there  can  he  no  root  so  great  as  that 
assumed. 

For  the  substitution  of  any  greater  number  will  make  all 
the  sums  still  greater,  and  so  wdll  carry  the  last  sum,  or  i^^*, 
still  further  from  zero. 

III.  If  the  sums  are  alternately  positive  and  nega- 
tive, the  value  of  x  we  employ  is  less  than  any  root. 

IV.  //  two  values  of  x  give  different  signs  to  Fx,  there 
must  he  one  or  some  odd  numher  of  roots  hetween  these 
values  (compare  §  345). 


GENERAL    THEORY   OF  EQUATIONS,  453 

V.  Two  values  of  x  which  lead  to  the  same  sign  of  Fx 
iiicliule  either  no  roots  or  an  even  number  of  roots  be- 
tween them. 

Let  us  take  as  a  first  example  the  equation 

2;3  _  72:  -h  7  =  0. 

Let  us  first  assume  2;  =  4.     We  compute  as  follows  : 

Coefficients,  1  0  —7  +7 

Products,  4  16  36 

Sums,  +1  T^       T43 

So  F  {i)  rrz  +43,  and  as  all  the  coefficients  are  positive, 
there  can  be  no  root  as  great  as  4. 

Putting  2;  =  —  4,  the  sums,  including  the  first  coefficient 
1,  are  1,  —4,  +9,  —29.  These  being  alternately  positive  and 
negative,  there  is  no  root  so  small  as  — 4. 

Substituting  all  integers  between  —4  and  -j-4,  we  find 

i^(-4)  =  -29,  F{0)  :=:  +   7, 

F{-^)=  +  1,  F{1)  =  +  1, 

F{-2)  =  4-13,  F{2)  =  +  1, 

F{—1)  =  +13,  i^(3)  =  +13. 

If  we  draw  the  curve  corresponding  to  these  values  (§  345), 
we  shall  find  one  root  between  —3  and  —4,  and  very  near 
—3.05,  and  the  curve  will  dip  below  the  base  line  between  +1 
and  +2,  showing  that  there  are  two  roots  between  these  num- 
bers ;  that  is,  there  are  two  roots  of  the  form  l-\-h,  li  being  a 
positive  fraction.  Transforming  the  equation  to  one  in  A, 
by  putting  1  +  A  for  .r,  we  find  the  equation  in  h  to  be 

7^3  ^  3/^2  _  47^  ^  1   :3::   0.  (1) 

Substituting  7^  =  0.2,  0.4,  0.6,  0.8,  we  find  that  there  is 
one  root  between  0.3  and  0.4,  and  one  between  0.6  and  0.7. 
Let  us  begin  with  the  latter. 

If  in  the  last  equation  we  put  7^==0.6  +  ^',  we  find  the 
transformed  equation  in  li^  to  be 

Fli!  =  h'^  +  4.87/2  ^  0.687/  -  0.104  =  0.  (2) 

If  we  substitute  different  values  of  h'  hi  this  equation,  wo 
29 


451  GENtJUAL    THEORY   OF  EQUATIONS. 

shall  find  that  it  must  exceed  .09,  and  as  it  must  be  less  than 
0.1,  we  conclude  that  9  is  the  figure  sought,  and  put 
h'  =z  .09  +  h". 

Transforming  the  equation  (2),  we  find  the  equation  in  h" 

to  be 

h"^  +  o.07A"2  +  1.5683//'  -  0.003191  =  0.  (3) 

Since  h"  is  necessarily  less  than  0.01,  its  first  digit,  which 
is  all  we  want,  is  easily  found,  because  the  two  first  terms  of 
the  equation  are  very  small  compared  with  the  third.  So  we 
simply  divide  .003191  by  1.5683,  and  find  that  .002  is  the  re- 
quired digit  of  h".     We  now  put 

h"  =  .002  +  h'", 
and  transform  again.     The  resulting  equation  for  h'"  is 

jr^  +  5.0767z'"2  +  1.588592y^'"  -  0.000034112  =  0.     (4) 

The  digits  of  x,  h,  Ji',  and  h"  which  we  have  found  show 
the  true  value  of  a;  to  be  , 

X  =  1.692  +  h'". 

By  continuing  this  process,  as  many  figures  as  we  please 
may  be  found.  But,  after  a  certain  point,  the  operation  may 
be  abbreviated  by  cutting  off  the  last  figures  in  the  coefficients 
of  the  powers  of  h. 

The  work,  so  far  as  we  have  performed  it,  may  be  arranged 
in  the  following  form  (see  next  page). 

The  numbers  under  the  double  lines  are  the  coefficients  of 
the  powers  of  h,  h',  h",  etc.  It  will  be  seen  that  for  each  digit 
we  add  to  the  root,  we  add  one  digit  to  the  coefficient  of  li^, 
two  to  that  of  h,  and  three  to  the  absolute  term.  We  have 
thus  extended  the  latter  to  nine  places  of  decimals,  which,  in 
most  cases,  will  give  nine  figures  of  the  root  correctly.  If  this 
is  all  we  need,  we  add  no  more  decimals,  but  cut  off  one  from 
the  coefficient  of  h,  two  from  that  of  h^,  and  so  on  for  each 
decimal  we  add  to  the  root. 

We  shall  find  the  next  figure  after  1.692  to  be  zero  ;  so  we 
cut  off  the  figures  without  making  any  change  in  the  coeffi- 
cients. The  next  following  is  2,  so  we  cut  off  again  for  it,  and 
multiply  as  shown  in  the  following  continuation  of  the  process : 


GENERAL    THEORY  OF  EQUATIONS. 


455 


0 

-7 

+1 

+  1 

+1 

-6 

+1 

+2 

+  2 

-4.00 

+  1 

+  2.16 

+  3.0 

-1.84 

+  .6 

+  2.52 

+  3.6 

+  0.6800 

.6 

+  0.4401 

4.2 

+  1.1201 

6 

+   .4482 

+  4.80 

+  1.568300 

9 

10144 

4.89 

+  1.578444 

9 

10148 

4.98 
9 

+  1.58859^ 

+  5.070 

2 

5.072 

2 

5.074 

2 

+  7 
-6 


1.692 


+  1.000 
-1.104 

-  ,104000 
+  .100809 

-  .003191000 
+  .003156888 

-34112 


+  5.076 


CONTINUATION  OF  FROCESS. 


+ 15.076 


+  1.5885|9!2 
1 

-34112 
31774 

1.5887 
1 

-2338 
1589 

1.5|8|8|8 

—  749 
636 

—113 
111 

-2 

I  021471 


It  will  be  seen  that  from  this  point  we  make  no  use  of  the 
coefficient  1  of /^^,  and  only  with  the  second  decimal  do  we  use 
the  coefficient  of  JiK  After  that,  the  remaining  four  figures 
are  obtained  by  pure  division. 

There  is  one  thing,  however,  which  a  computer  should 
always  attend  to  in  multiplying  a  number  from  which  he  has 
cut  off  figures  in  this  way,  namely : 

Always  carry  to  the  product  the  ninnher  ichich  would 
have  been  carried  if  the  figures  had  not  hceri  cut  off,  and 


456 


GENERAL    THEORY   OF  EQUATIONS. 


increase  it  hjj  1  if  the  figure  following  the  one  carried 
would  have  been  5  or  greater. 

For  instance,  we  had  to  multiply  by  7  the  number  151888. 
If  we  entirely  omit  the  figures  cut  oif,  the  result  would  be  105. 
But  the  correct  result  is  1111216;  we  therefore  take  111  in- 
stead of  105. 

Again,  in  the  operation  preceding,  we  had  to  multiply 
158|88  by  4.  The  true  product  is  635|52.  But,  instead  of 
using  the  figures  635,  we  use  636,  because  the  former  is  too 
small  by  |52,  and  the  latter  too  great  by  |48,  and  therefore  the 
nearer  the  truth.  For  the  same  reason,  in  multiplying  1.58818 
by  1,  we  called  the  result  1589. 

Joining  all  the  figures  computed,  we  find  the  root  sought 
to  be  1.692021471. 

Let  us  now  find  the  negative  root,  which  we  have  found  to 
lie  between  — 3  and  — 4.  Owing  to  the  inconvenience  of 
using  negative  digits,  and  thus  having  to  change  the  sign  of 
every  number  we  multiply,  we  transform  the  equation  into  one 
having  an  equal  positive  root  by  changing  the  signs  of  the 
alternate  terms.     The  equation  then  is  2;^  —  Ix  —  7  =:  0. 

The  work,  so  far  as  it  is  necessary  to  carry  it,  is  now  ar= 
rammed  as  follows: 


0 
3^ 
3 
3 

6 

o 
o 

4 

9.04 
_4 

9.08 
_^ 

9120 

8 

9.128 

8 

9.136 

^ 

(9l|44 


-7 

2 
18 


-7  'I  3.0489173395 
6 


20.0000 

.8616 
20.3616 

.3632 

20?724800 
73024 

20.7978-M 
73088 

20.87091 12 

823 1 0 

20.87914|2 

823 

20T8873J7 
9 

20|.8|8|7|5 


-1.000000 
814464 

-0.185536000 
.166882592 

-     .19153408 

18791228 

-862180 

2088':  5 

^53305 

146213 

-7092 

6266 


627 
-199 

188 
-11 


OENEUAL    THEORY  OF  EQUATIONS.  4^7 

The  negative  root  of  the  equation  is  therefore 
—  3.0489173395. 

EXE  RCI  SES. 

Find  the  roots  of  the  following  equations: 

1.  a;s  _  3^2  ^  1  _  0  (3  real  roots). 

2.  x^  —  Zx  -\-l  —  0  (3  real  roots). 

3.  x^  —  4^^  -^  2  =2  0  {%  positive  roots). 

4.  ^;2   +   ,^;  _   1    —    0. 

5.  Prove  that  when  we  change  the  algebraic  signs  of  the 
alternate  coefficients  of  an  equation,  the  sign  of  the  root  will 
be  changed. 

372.  The  preceding  method  may  be  applied  without 
change  to  the  solution  of  numerical  quadratic  equations,  and 
to  the  extraction  of  square  and  cube  roots.  In  fact,  the  square 
root  of  a  number  ^  is  a  root  of  the  equation  x^  —  ^  =  0,  or 
x^  -^  Ox  —  n  =z  0,  and  the  cube  root  is  a  root  of  the  equation 
a^  -\-0x^  -\-()x  —  n  =  0, 

Ex.  I.  To  compute  a/2. 

10  -2  1  1.4142135a 

1  1  ■ 


1  -1.00 

1_  .96 

2:0 
0.4 


-.0400 
281 


24  -11900 

4                 -        -  11296 

=  -60400 

2.80  56564 

2.81 

— J-  -1008 

2.820  849 


1  -159 

2.824  JM 

-18 


2.8280  IZ 
2  1 

2.8282 
2 


2|.8|2|8|4 


458  GENERAL    THEORY   OF   EQUATIONS. 

Ex.  2.  To  compute  the  cube  root  of  9842036. 


0 

0 

2 

4 

2 

4 

2 

8 

4 

1200 

2 

61 

60 

1261 

1 

62 

.— -- 

61 

132300 

1 

2536 

62 

134836 

1 

2552 

630 

137388.00 

4 

192.69 

634 

137580.69 

4 
638 

192.78 

137773.47 

1.93 

642.0 
.3 

1317|7175|4 

642.3 

3 

642.6 

3 

-9842036  I  214^30303242 
8  ■ 

-1842 
1261 


-581036 
539344 


41692000 

41274207 

-417793 

413326 

4467 

4133 


276 

58 
55 

8 


:64^i^ 


APPENDIX, 


SUPPLEMENTARY    EXERCISES. 


Note.  The  following  additional  exercises  and  problems  are  of  the 
same  general  character  with  those  in  the  body  of  the  book.  They  are 
partly  original,  and  partly  selected  from  the  best  recent  German  col- 
lections of  problems.  They  are  arranged  under  the  section  numbers 
to  which  they  pertain,  so  that  the  teacher,  on  arriving  at  those  sections, 
will  be  able  to  select  as  many  of  them  as  he  deems  necessary  for  the 
drill  of  his  class. 


SUPPLEMENTARY  EXERCISES. 


Algebraic  Addition  and  Subtraction. 

§15. 

Supposing  one  to  start  from  a  certain  point  on  the  scale 
of  numbers,  and  then  move  over  positive  and  negative  spaces 
as  follows,  it  is  required  to  find  his  stopping-point  in  each 
of  the  following  cases: 

1.  Starts  from  +  4,  and  moves  through  +  2  —  3  +  9  —  7 

—  2  units. 

2.  Starts  from  +  9,  and  moves  through  —  1  —  6  —  9  +  5 
+  8  units. 

3.  Starts  from  —  1,  and  moves  through  +  2  —  3  +  4  —  5 
+  6  units. 

4.  Starts  from  —  8,  and  moves  through  —  1  +  3  —  5  +  7 

—  9  units. 

5.  Starts  from  —  12,  and  moves  through  —9-6  +  8  +  5 
+  8  units. 

§31. 

I.  How  far  is  A  from  B  (positively  or  negatively)  when 
they  have  severally  made  the  following  motions  from  the  same 
point  on  the  scale  of  numbers: 

, A .  , B . 

a.  -2-3-5  +  7.  +1  +  2  +  3  +  4  +  5. 

2>.    -5  +  5-6  +  6.  +5  +  6-2-4  +  12. 

0,    -2  +  7  +  8  +  9  +  10.       -7-3  +  4-5-6. 

a,    _i-2  +  6-2-l.         +3  +  4  +  5-8-3.    Ans.-l. 


462  ALGEBRAIC  ADDITION 

2.  What  is  the  meaning  of  the  following  expressions  : 

That  man  is  —  6  years  older  than  his  wife  ? 
Kichmond  is  —  70  miles  north  of  Washington  ? 
You  are  —  3  inches  taller  than  your  brother  ? 

3.  The  Autocrat  of  the  Breakfast  Table  tells  of  a  Parson 
Turrel  who,  dying  in  the  last  century,  bequeathed  a  noted 
chair  to  the  oldest  member  of  the  Senior  class  in  Harvard 
College,  which  was  to  be  passed  down  from  class  to  class 
indefinitely.  The  first  Senior  who  got  it  was  to  pay  5  crowns, 
but  each  succeeding  one  was  to  get  it  at  a  price  1  crown  less 
than  that  paid  by  his  predecessor.  How  would  the  require- 
ment of  the  will  work  at  the  end  of  7  and  of  100  years? 

§34. 

I.  Find  the  value  oi  a  —  h  and  of  b  —  a  when  a  and  b  have 
the  following  sets  of  values  : 

(1)  (2)  (3)  (4)  (5)  (6)  (7)  (8)  (9)  (10) 
a=  +2,  +7,  -9,  -5,  -17,  +  8,  -33,  -18,  +12,  +22 
b=   -3,  -9,  -3,  +8,  -29,  +14,  +13,  -19,  -12,  -22 


a-b=  +5    

b—a—   —5    

2.  Compute  the  values  of  1  +  3^  and  of  1  —  3:^;  for  the 
following  11  values  of  x  : 

:^  =-  5,  -  4,  -  3,  -  2,  -  1,  0,  +  1,  +  2,  +  3,  +  4,  +  5. 

3.  Compute  the  values  of  a  -\-2b  and  ot  a  —  2b  for  each 
of  the  10  sets  of  values  of  a  and  b  in  Ex.  1. 

§56. 

1.  How  much  is  a  -\-  2x  greater  than   a  —  3x,  and  vice 
versa  9 

2.  How  much  \s  a  —  b  greater  than  b  —  a? 

3.  How  much  is  0  greater  than  a  —  2b? 

4.  How  much  is  0  greater  than  —  x?     Than  -\-  x? 


AND  SUBTliACTION,  463 

5.  A  party  of  9  boys  were  formed  into  a  solid  square  of 
3  rows,  with  3  boys  in  each  row.     The  rear  left- 
hand  boy  B  was  t  inches  tall.     Every  other  boy 

was  X  inches  taller  than  the  boy  next  behind  him, 
and  2/  inches  shorter  than  the  boy  on  his  left.    Ex- 
press the  height  of  each  boy,  and  the  sum  of  the  heights  of 
all  the  boys. 

6.  During  six  successive  days  a  man  earned  m  cents  more 
every  day  than  he  did  the  day  before,  and  paid  out  n  cents 
less.  On  the  first  day  his  earnings  were  h  cents,  and  his  pay- 
ments h  cents.  How  much  had  he  left  at  the  end  of  the 
sixth  day? 

7.  Of  two  travellers,  X  went  east  h  miles  and  then  returned 
h  miles  toward  the  west ;  Y  went  west  x  miles  and  then 
returned  y  miles  toward  the  east.  If  they  started  together, 
how  far  was  X  east  of  Y  when  they  stopped  ?  How  far  was 
Y  east  of  X  ? 

8.  There  were  three  travellers  on  the  same  road,  B  being 
X  miles  west  of  C,  and  C  y  miles  west  of  A.  A  went  m  miles 
toward  the  east ;  B  went  twice  as  far  as  that  toward  the  east; 
and  C  went  4??i  miles  toward  the  west.  How  far  was  each 
west  of  the  two  others  when  they  stopped  ? 

9.  Of  two  men,  A  and  B,  A  had  a  dollars  and  B  had  x 
dollars  on  Monday  morning.  On  Monday  evening  A  paid  B 
d  dollars,  and  B  returned  y  dollars  of  this  to  A.  Each  fol- 
lowing evening  during  the  week  A  paid  B  g  dollars  less 
than  before,  and  B  returned  A  z  dollars  less  than  he  did  the 
evening  before.  How  much  had  each  on  each  morning  from 
Tuesday  to  Saturday? 

10.  Four  casks,  marked  A,  B,  C  and  D,  each  containing  r 
gallons  of  water,  stood  at  the  corners  of  a  square.  Then  m 
gallons  were  poured  out  of  A  into  B,  71  gallons  out  of  B  into 
C,  p  gallons  out  of  0  into  D,  and  q  gallons  out  of  D  into  A. 
How  much  was  then  in  each  cask  ?  Prove  the  result  by 
showing  that  the  sum  of  the  quantities  in  all  the  casks  is  4r. 

II.  The  same  four  casks  at  first  contained  a,  h,  c  and  d 
gallons  respectively.  Then  x  gallons  were  poured  out  of  B 
into  A.     Then  a  quantity  equal  to  what  was  left  in  B  was 


464  ALGEBRAIC  ADDITION. 

poured  from  C  into  B  ;  a  quantity  equal  to  wliat  was  left  in  C 
was  poured  from  D  into  0  ;  and,  finally,  a  quantity  equal  to 
what  was  left  in  D  was  poured  from  A  into  D.  How  much 
was  then  left  in  each  cask  ?    Prove  as  before. 

12.  Three  traders,  A,  B  and  0,  had  a,  1)  and  c  dollars 
respectively.  A  bought  c  dollars'  worth  of  goods  from  B  ;  B, 
a  dollars'  worth  from  C  ;  and  C  bought  I  dollars'  worth  from 
A.  When  each  paid  the  other  for  the  goods,  how  much 
money  had  each  left  ?  What  was  the  sum-total  of  money 
possessed  by  the  three  ? 

13.  Given  a  quadrangle  the  lengths  of  whose  sides  are  a, 
l,  c  and  d  respectively.  Enough 
of  the  side  b  is  cut  off  and  added 
to  a  to  double  the  latter;  the  re- 
mainder of  I  is  then  doubled  by 
cutting  off  from  c ;  and  the  re- 
mainder of  c  is  doubled  by  cutting 

off  from  d.     How  long  will  each  side  then  be  ? 

14.  Of  two  men  starting  out  from  the  same  point,  A 
walked  m  miles  west  the  first  day,  and  h  miles  more  each  fol- 
lowing day  than  he  did  the  day  before  ;  B  walked  p  miles 
west  the  first  day,  and  x  miles  less  each  day  than  ho  did  the 
da^  before.  How  far  was  A  west  of  B,  and  how  far  was  B 
west  of  A,  at  the  end  of  the  first,  second,  third  and  fourth 
days  respectively  ? 

15.  If,  on  this  line,  we  suppose  the  point  B  to  be  at  the 

East.  ^ ? ^^West. 

dislance  h  west  of  A,  and  C  to  be  at  the  distance  c  west  of  A, 
then,  ill  alirebraic  language: 

How  fjir  is  A  west  of  B  ? 

How  far  is  A  west  of  0  ? 

How  fjir  is  0  west  of  B  ? 

How  for  is  B  west  of  C  ? 

How  far  is  the  middle  point  between  B  and  0  west  of  A  ? 

How  far  is  the  middle  point  between  C  and  A  west  of  B  ? 


CLEABING  OF  PARENTHESES,  465 

How  far  is  the  middle  point  between  A  and  B  wesb  of  0  ? 
What  is  the  algebraic  sum  of  these  last  three  distances  ? 

Note.  Should  the  student  find  any  difficulty  in  this  or  the  next 
question,  he  should  begin  by  expressing  the  distances  a  and  h  in  num- 
bers, and  noticing  the  processes  by  which  the  measures  are  found. 

1 6.  The  three  points  A,  B  and  C  are  at  the  respective 

M A B C 

I  ill 

distances  a,  h  and  c  west  of  a  fourth  point,  M.    Express  alge- 
braically the  three  distances 

B  west  of  A ;  A  west  of  C  ;  0  west  of  B, 

and  take  their  sum.     Express  also  the  distances 

A  west  of  the  middle  point  between  M  and  B, 
B  "  "  "  M  and  C, 

C  '^  "  "  MandA, 

and  find  the  sum  of  these  three  distances.     Then  express 

A  west  of  the  middle  point  between  B  and  0, 
B  "  "  "  C  and  A, 

C  "  "  "  A  and  B, 

and  find  the  algebraic  sum  of  tlie  three  distances.      Express 

also  the  three  mutual  distances  !)ct\veeu  tbe  middle  points  of 

lines  AB,  BO  and  CA  respectively — that  is  : 

Mid.  point  betw.  A  and  B  west  of  mid.  point  betw.  B  and  0, 

etc.  etc.  etc. 

§61. 

Clear  the  following  expressions  of  parentheses,  and  com- 
bine the  terms  by  addition: 

1.  3  m  —  [A  —  2m  —  {Ji -\- m)  —  (h  —  m)\ 

2.  (ct  —  y)  ~  (a  '\- V)  -^  (a  —  m)  —  {a  —  m). 

3.  (a  +  h)-{a-^h)^  \_{a  ^  h)  ^{a  +  ^)]. 

4.  Vi  —  {37^  —  [4A  —  {bh  —  m)  +  m]  +  2m}. 

5.  3c-  'M-(2d-'dc)  +  [-  [c-  d)  -  (3c  +  2^?)]' 

6.  A:h  -  In  -  (4.h  +  In)  -  [3A  +  (4m  -  a)  -  (5m  +  h)]. 


466  MULTIPLICATION, 

7.  x-\-  {x  —  a—  {2a  —  2x)  +  [a  —  {a  ~  x)] }, 

8.  6a+  {6a  -~  2x+[4a  -  Sx  -  {3a  -  4.x)]\. 

9.  {a  +  b  -  c)  +  {a-  b  +  c)+{-  a  +  b  +  c)-{a  +  b  +  c). 

10.  a  +  dx—  {2a  +  2x)  —  {da  +  x)  —  [a  —  {a  —  x)]. 

11.  b  +  [-2b-dc-{dc-b)-3b]. 

12.  -[-  (3w-27i)  +  (2m-3/i)]  +  [{5m-A7i)-{3?i~4:m)]. 

Multiplication  and  Addition. 

§74. 
■  Clear  the  following  expressions  of  parentheses: 

1.  b  {  a{c  —  x)  -\-b{c-\-x)  -{-  ax[b  ~  c{x  —  a)]}, 

2.  m  [x  —  n  {b  —  y)  -\- b  {n -\-  y)  -\-  y  {n  +  b)\ . 

3.  a?i  [an  (1  —  a7i)  -\-  a'^n''  (1  —  an)\, 

4.  h{l-Vh[l  +  h{l  +  li)^\. 

5.  x{l-x\l-x{l-x)']}. 

6.  X  {2)  -{-  x\_q~\-x  (r  +  ^)]  }, 

7.  a;  {;?  —  ir  [^  —  a;  (?' —  a;)]}. 

9.   a  I  [(a:c  —  Z>)  ic  —  c]  2:  —  6?}. 

10.  i>i[(;?^+/).'?^+/]a:+y}. 

11.  {  [{7nx  —  nt")  X  —  m^]  .t  —  ^m^  }  mx, 

12.  [a'  (^  -  c)  +  b'  {c  -  «)  +c=^  {a  -  b)]  abc, 

13.  m  [a'  {x  +  y)  +  b'  {x-y)-  x{a'  +  ^'^J  +  {a'  -  b')  ym. 

14.  a  {«  —  ^  [rt  —  c  («^  —  d)]}, 

.15.  «  |aZ>  — c  [a^^— c^  (a''^  —  (i')]}. 

16.  (^  +  a;)(^-.v)  +  («^-a;)(^+2/)- 

17.  {m -\- n)  {x -\- y)  —  {m  —  n)  {x  —  y), 

(§  76.) 

Arrange  the  following  expressions  according  to  powers 
of  x: 

1.  {x'-x'+l)a'  +  {x'--'X+l)a'+a\ 

2.  1  -{-X  —  x^  "-x^  —  a{l  -{-X  —  X*)  -\-a'^{l  —  x)  —  a^» 

3.  7na^—-na^{x  —  l)--ma''(^'^— a:  +  1)"~  ^i<:i^{^'— ^''+i^  —  1). 

4.  «  —  a;  { ^  —  :?;  [c  —  c?;  (^  —  a:)]  } . 


DIVISION,  467 

5.  { {ax  —  h)  {ax  -\-h)  —  {bx  —  a)  {hx  +  a)  ]  {a^  +  h'')x. 

6.  I  {mx  -\-  ay  -[-  (mx  —  aY]  (mV  —  a^), 

7.  { {inx  +  ccY  —  [mx  —  ay  \  (m  —  nx)  {m  +  fix), 

8.  a  +  X  {b  +  X  [c  +  X  {d  +  x)]\ 

~  a  \x  -^^  I  \x -\-  c  {d  -^  x)\\. 

§80. 

Write  out  the  results  of  the  following  powers  and  pro- 
duets  on  sight: 

I.   {ax-\-lyy.  2.  {ax  —  hyy. 

3.  {ay  +  lxy.  4.  {ay  —  bxy. 

5.  {ax-^2byy.  6.  {ax-2byy. 

7.  {ax  +  dbyy.  8.  {ax-dbyy. 

9.   (m  +  ny  X.  10.  (771  -j-  ^^)'^  ^. 

II.  ir(a;+.v)*-  12.  ic  (:?;  —  ?/)'. 

13.  a(a:  — ,?/)  (:r  +  ^).  14.  ax''  {a  ~  x)  {a-}-x). 

15.  7??/^  (2m  +  n)  {2m  —  7^).       16.  mW  (3m  +  ^?.)  (3m  —  n). 

17.  (^  +  by  +  {a-  by.        18.  (6j  +  by  -{a-  by. 

Form  the  values  of  the  following  quantities,  and  arrange 
according  to  powers  of  x,  y  and  z\ 

19.  {ax-\-byy -\-{bx  —  ayy. 

20.  {ax  -\-  byy  ~  {bx  —  ayy 

21.  {2mx  —  nyy  -\-  {mx  —  2nyy. 

22.  {2mx  —  nyy  —  {mx  —  2nyy, 

23.  {x  -\-  ny)  {x  —  7iy)  {x^  —  n'y^) 

—  (y  +  nx)  {y  —  nx)  {y^  —  n^x^). 

24.  {ax  -\-  by  -{-  cz)  {ax  +  by  —  cz) 

{ax  —  by  -\-  cz)  {ax  —  by  —  cz). 

Division. 

§(85.) 

1.  Mx^'bc^  -^2ax^b. 

2.  24«':r"?/ -V- 6fl^'^V- 

3.  12a^icy -^  4a'icV. 

4.  a'^o;  (c  +  ^ )  -^  ax{c-\-d). 

5.  a  C^-y)  -^a(ic~-2/). 


468  FACTORING, 

6.  'bc{x  +  y)-^h{x-^y). 

7.  x^'y  {a  —  h)  -^  X  {a  —  Z>), 

8.  10.T^  (a-V)-^  ^x^  {a'\'V)-^x. 

10.  10  {a  +  Vf  -  15  {a  +  ^)^  ~  10  (^  +  Z>)  ~  5  (a  +  Z>). 

12.  (a+a;r(a+2/r-(«+^r(^+2/r. 

13.   —V^oTl'^-^^a^h'^, 

Factoring. 

§89. 

Factor  the  following  expressions: 

2,  X  —  ax^  +  ^^^-  3-  —  ^'  +  m^?i  —  m^n^. 

4.  «m  —  «'^/i^  +  a%^  5.  c"Z>'"  —  c^'^Jf'. 

6.  —  abc-\-if)fabc,  7.  ft"^^^**  —  a^:c" "  \ 

8.  ahVy"^  —  a'lfx^y,  9.  m/z-^jf?^  —  S7n^7i^p. 

10.  2a"ic'"  -  7a'":c^  11.  8a^c  -  12aWc\ 

§91. 

In  the  following  exercises,  first  take  out  all  monomial 
factors  common  to  the  several  terms,  as  in  §  89,  and  factor 
the  remaining  terms  by  the  rules: 

1.  a^  —  aV^.  2.  mW  —  w*. 

3.  ^ififx  —  ^yi^x.  4.  mx^  —  m. 


5. 

aV  -  a\ 

6. 

ax^  —  ax. 

7. 

rrfx^  —  m*x. 

8. 

9X'    -   4:X\ 

9. 

aV  -  4:a'x\ 

xo. 

7n''y^  —  7/iy. 

II. 

16my  -  2bmY' 

12. 

4.9aV  -  ma'x. 

13. 

am^  -a^  +  ^a^h  -  ab\ 

14. 

a.V  -  4bV  +  4hcx^  —  c'x\ 

15. 

«ic'"  -  4a'2:"  +  ^a\ 

16. 

a'h  -  4.a'F  +  4.ah\ 

17. 

4yx'  -  12xY  +  9x'y\ 

18. 

^xY  +  12^:^'  +  ^^Y' 

§93. 

1.  2  {x'  +y*  +  z*)  -4:  {xY  +  y'z'  +  zV). 

2.  a'  +  16b'  +  c'-  Sa'F  -  SFc'  -  2cV. 


FAGTOUING.  469 

3.  2  (xY  +  y'^z'  +  z'x')  -x'  -  f  -  z\ 

4.  8a^Z>V+  Z%Ve  +  8cV  -  ^*  -  16^^  -  16c*. 

§94. 

In  the  following,  begin  by  removing  all  factors  common 
to  the  two  terms  as  in  §  89: 

1.  «^  +  4a V  +  4a V.     A71S,  a'  {a  +  2^;')'. 

2.  a^  —  aV.     Ans,  a''  {a  -\-  x)  {a  —  x)  {a*  +  ^^^^  +  ^*)- 


3.  «V  —  x". 

4.  ay'  —  ay  . 

5.  (a  -  S)^  -  c\ 

6.  («  +  ^)^  -  c^\ 

7.  a;'  (a;  -  yY  -  a;^ 

8.  :c*  +  8a:^/^ 

9.  {a  +  by  +  a\ 

10.  a{a  +  ly  +  a\ 

II.  a;'  +  /. 

12.  a'+  aZ>'. 

13.  a"  +  64m V. 

14.  m'  +  Um'x'. 

IS.  a;=  +  ^'. 

16.  a' +  8. 

17.  «'  +  216. 

18.  64:^''  +  125cl 

19.  a;^  +  /- 

20.  ic'  —  a'. 

21.  Sa'  -  27b\ 

22.  64m'  -  M\ 

23.    .-K'  +  l. 

24.  64a'  +  Z^^ 

25.  a'  +  ab'. 

26.  a  -  27a\ 

27.  ab'-b\ 

28.  a'  -  243. 

29.  32a'°  +  l. 

30.  16a'  —  a. 

31.  3a;' +  16. 

*32.  27a^  +  8a2:^ 

33-   (2^  +  2/)'-  (a;- 

-yy 

34.   {x  +  f/)'  +  (^  -  2^)'. 

35-  {x  +  yy-{x~ 

-yy 

2,6,  1  —  a^  +  2aa;  —  x". 

Factoring  Trinomials. 

A  trinomial  of  the  form 

x' 

-\-  ax  -\-h 

can  always  be  factored  when  we  can  find  two  numbers  whose 
sum  is  a  and  whose  product  is  Z>.  For  if  m  and  n  are  these 
numbers,  the  trinomial  is 

x^  -\-  {m  -\-  n)  X  -{-  mn, 

which  is  equal  to 

(x  -\-  m)  (x  -\-  n).  ♦ 


470  FACTO  BIN  Q. 

Factor: 

1.  x"  -{-  {a-\-  h)  X  +  ah.     Arts,  {x  -{-  a)  {x -{-  h), 

2.  y'+^  +  2.     Ans.  {^  +  l)(y  +  2). 

3.  2/^  +  4«^  +  3.  4.  cr^  +  5.T  +  4. 
5.  n"  +  5yi  +  6.  6.  t^'^  +  i5n  +  8. 
7.  a'  +  7«^  +  10.  8.  a'  +  8a  +  12. 
9.  m'  +  7m  +  12.  10.  m'*  +  8m  +  15. 

II.  x'  +  Ix"  +  10a;.  12.  y'  -\-  6y'  +  8y. 

13.  x'  +  7x'  +  12a;^  14.  a' +  Sa'  +  15a'. 

15.  x'  -f  19^;^  +  88.  16.  a'  +  12a'  +  35. 

17.  x'""  +  9^:^  +  20.  18.  y'""  +  by'""  +  6. 

19.  ir''  +  (^  ""  ^^)  ^  ""  ^^-  ^^s.  {x  +  m)  (a;  —  7i), 

From  this  last  example  it  is  seen  that  when  the  quantities 
m  and  n  have  opposite  signs  the  last  term  of  the  trinomial 
will  be  negative^  while  the  middle  term  will  have  the  sign  of 
the  greater  of  those  quantities,  being  equal  to  their  algebraic 
sum  or  numerical  difference. 

20.  x^  —  X  —  6,     Ans.  {x  —  3)  (x  -{-  2). 

21.  x'  -x'-  12.  22.  y*  -  2y'  -  15. 
23.  a'  +  a  —  30.  24.  a^  —a  —  30. 
25.  m*  +  2m  —  8.  26.  m"^  —  2m  —  8. 
27.  7i'  -  dn'  -  40.  28.  m'  +  3m'  -  40. 
29.  a;^  +  {2a  —  db)  x  —  6ab,       30.  x''  —  dax  —  4td'. 
31.  a;*  +  ao;'  -  6a\  32.  a:''*  -  Ux''  -  12b\ 

If  the  quantities  m  and  7i  are  both  negative,  the  sum 
m  +  n  will  be  negative  and  the  product  positive,  because 

{x  —  m)  (x  —  n)  =  x^  —  {m  -\-  n)  x  -\-  mn. 

ZZ-  x"  -  {a-^b)x  +  ab.  34.  y'  -  oy  +  6. 

35.  f-^  +  ^'  36.  x-" - nx  + 10. 

37.  x"  -  IZx  +  40.  38.  x"  -  8.T  +  15. 

39.  ax"  —  dax  +  2a,  40.  ax^  —  Qax""  -{-  Sa, 

41.  7n^x^  —  bmx  -j-  4.  42.  m.Y  —  5ma:  +  4. 

43.  ml'?;^  —  dmx  +  2.  44-  ^^^^  —  4:ma:  +  4. 

45.  aV  —  la^'x  -\-  12a,  46.  mV  —  7m V  +  12m*. 

47.  Tiy  -  7/^y  +  10n\  48.  ry  -  7r'?/'  +  12ry. 


FACTORING.  471 

In  the  following  exercises  trinomials  of  all  the  preceding 
classes  are  contained: 

I.  x^  +  10^  +  24.  •     2.  x^  -Qx  +  8. 

3.  x''-\-^x-  20.  4.  x'  +  3:r'  +  2. 

5.  a;'  -  7:?:"  +  12.  6.  :cy  -  ^Ixy  +  26. 

7.  a'c'  -  Uabc  +  396'^  8.  a'b''  -  Ma'hx  +  143:r^ 

9.  a'  -  12a  +  20.  10.  x'  +  5O2;  +  49. 

II.  a;' +  4a: -32.  12.  a^- 7a -18. 

13.  a'  +  a'  -  132.  14.  a'^^V  +  Wb^d"  -  22. 

15.  a*  +  17a'  -  390.  16.  a"  -  7a  -^  12. 

17.  x'  +  x-  72.  18.  a;^  -  12a;  +  27. 

19.  x"  —  39x  +  108.  20.  a;'  —  a;  —  12. 

21.  a;'*  -  7a;  -  60.  22.  15a;*  -  17a;^  +  4. 

23.  (a  +  by  -  lie  (a  +  b)  +  dOc\ 

24.  a;^  +  4a;  -  77.  25.  a;'  +  6a;  -  135. 
26.  a;'  — 14a; +  48.  27.  a;' +  12a;  +  35. 

Miscellaneous  Exercises  in  Factoring. 

I.  ax'  —  2bx  -\-  ex.  2.  ax  —  2^a;  +  dcy  —  ex  -\-  2y. 

3.  a^x  —  2eay  —  4:X  —  y  +  x.  4.  a^x  +  a^^'o;^  —  3a^"'a;'. 

5.  ax'—  2bx^~\-  ex\  6.  tx'  —  pqx^  -{-px"". 

7.  ex'  —  abx^  —  2y  -\-  Say'.      8.  aexy  +  20-^  —  dx'y*. 

9.  2c'x'y'  —  x^y"^  +  dx'y'.       10.  4:x'y  —  3a;^?/'^+  2x^y'. 

II.  (a;*  -  4). 

13.  {x'  -  81a;''). 

15.  o;^  -a;  +  i. 

17.  a^Z^'^  -  a^Z>^  +  ^. 

19.  ia'  -\-l  —  4:a. 

21.  a;'  +  2a;'y  +  xy'. 

23.  25aV  -  30aa;^.V  +  9a;y . 


28.  a' 


x' 

^5.-4- 

3y  ^  9^"- 

x'          a; 
6y'       18,«/' 

12. 

(o;'  -  9a;). 

14. 

4.x'  -12x'y  +  9y'. 

16. 

a'x'  -  y'. 

18. 

9a'  -  1. 

20. 

a;^^  +  a;*"  +  i. 

22. 

16a*^  -  1. 

24. 

12a*  -  36a'xy  +  27x'y\ 

26. 

a'        «    ,      1 
8        6^  "^  ISb'' 

472  FAGTOEING. 

29.  %4.a'h  -  na'V  +  b4.ah\     30.  x'*  -  y'\ 

31.  (4  — gij-  3--  mV  +  2mV  +  mV. 

33.  289a'^Z^^c'  +  102a;>^c'^J  +  9^V^. 

34.  121a'  -  2^Qa'¥  +  169Z>\    35..  98a''Z>^  -  b^ahx  +  8:c\ 

36.  16^;^  +  ^x'  +  ^^  37.  fK  -  2/". 

38.  d^a'h^e  -  bVc\  39.  j-«  +  l. 

40.  16a^  {^aF  -  lObc)  +  20c  {6c  -  4:aFy 

41.  —  —  Sax''.  42.  2^xy  —  Id^xyz". 

43.  (a  -  by  -  (a  +  by.  44.  (3«  -  5)"  -  («  +  2*)'- 

45-  ^^^^-  -  ^^y.  46.  («  +  5)=  -  K  -2«&  +  J^). 

47.  (a  -  5)'  -  (4a''  —  12ab  +  9b'}. 

48.  {a  +  if  -  (4a'  +  12ab  +  9b'). 

49-  i  -  2  +  J.  SO-  «'  -  2««/"  +  2/^ 

X  tl> 

51.  ^'^  -4a;^^^  +  4^^.  52.  a'^*'^  -  2^c&  +  c\ 

53.  a^'^—y^^  54.  16a'  —  4c'. 

57.  4a;'  -  4.x'  +  i?;\  58.  4.aV  -  {a'  +  b'  -  cy, 

59.  {2x  +  yy-^{x  +  yy. 

Products  of  Two  Binomials. 

We  have 

{a  +  i)  {x  +  y)=ax+bx  +  ay  +  by. 

Hence  a  polynomial  of  four  terms  may  sometimes  be  ex- 
pressed as  a  product  of  two  binomial  factors.  We  can  do  this 
when,  two  terms  of  the  polynomial  {ax -}- bx  for  example) 
being  divided  by  a  common  factor  {x),  and  the  two  remain- 
ing terms  by  a  common  factor  {y),  the  quotients  are  equal. 
We  can  thus  factor  the  following: 

1.  ax  —  bx  -\-ay  ■—  by,     Ans,  {a  —  b)  {x  -}-  y). 

2.  ax  -{-bx  —  ay  —  by,  3.  ax  --  bx  —  ay  -}-  by. 


DIVISION  BY  POLYNOMIALS.  473 

4.  nv"  +  mn  -{-71^  +  n,  5.  mn  —  m^  -\-  if  —  tfm, 

6.  l  +  a  +  a"  +  a\  t.  l-x-x'  +  x\ 

8.  l-\-x  —  x''  —  x\  9.  a'  +  fl^^c?:  +  ax"  +  a:'. 

10.  (a  —  ?i:r)  {a  +  ^za;)  —  (w  — -  ax)  {n  +  a^). 

11.  h^  -  Wx  +  hx'  -  dx\        12.  a^  +  «^  -  a^  -  a\ 

13.  m^  —  3m'  +  m'  —  3m'.       14.  m'  +  3m'  —  m'  —  3m. 

Division  by  Polynomials. 

§97. 

1.  a"  +  4:ax  +  4a;'  ~  a  +  2x. 

2.  6a'  -  6b'  -f-2a'^  -  2b\ 

3.  «^  -  da'b'  +  3a'b'  -  b'  -^  a'  -  da'b  +  Sab'  -  b\ 

4.  a'  -  9«'  +  27a  -  27  -f-  r?  -  3. 

5.  48a''  -  Hea'b  -  64:ab^  +  106b'  ~  2a  -  3b. 

6.  ^a'  +  a'+^a  +  i^ia+l. 

7.  dSa'b'  -  77a^b'  +  121a'b'  ~-  Sa'b  -  7a¥  +  llab\ 

8.  100«*  -  UOa'b  +  2d5aW  -  dOa'b'  ^  ba'  -  2a''Z^. 

9.  37a'Z>'  -  26a''Z>  +  3a'  -  Uab'  --  3«'*  -  6ab'+  2b\ 

10.  0^"  +  ^  +  x'^y  +  a:?/"*  +  ,^"*  +  ^  -h  :z;"*  +  y*". 

11.  a^  +  a^'^'W''  +  Z>*»  -^  a^^  +  a"Z>"  +  b^, 

12.  10a'  -  27a'^  +  3Wb'  -  18«Z>'  -  8^'  -^  2a'  -  3fl^  +  UK 

13.  4a:;l  —  3o(^y^  —  y  -7-  x^  —  «/i 

14.  8al  -—  6a^  -\-  a^  --■  2a^  —  a^. 

15.  9a-2  +  12a-^^-4-^3a-^+2. 

16.  4a;  -  10.Tt  -  62:«;i  -  30^^:1  -f-  2:^:*  +  5. 

17.  a;^y~^+ ^~y  ^  ^~V  +  ^^~^' 

18.  x^""  —  f"" -^  x""  -  «/". 

19.  4:?;^  +  6  -  3bx'  +  58a;*  -  70a;'  -  23a;  -f-  6a;'  -  5a;  +  2  -  7a;'. 

^       19    2  2a^x       a'        .  a' 

20.  a;*  —  —  a  a;'  +  -^  +  ^  -^  ^'  —  2aa;  +  - . 

21.  {a'  -  2ab  +  b'  -  c')  {a  +  b  +  c)  -^  a  -  b  -  c. 

22.  {ax  +  byY  +  {ay  -  bxy  -^  a'  +  b\ 

23.  12a;'  -  14a;'  -  llo:'  +  19a;  -  6  —  3a;'  -  5a;  +  2. 

24.  4:a'F  -  3b\  +  llaF  +  12a'  -^  34  a'^  -^  ^'  +  6a'  -  hah. 

25.  6aZ>c'  -  9Z>'6''  +  ^a'Z^'  -  a'c'  -=-  3^c  -  ac  +  2ab. 

26.  2Z>c;  -  1  +  a'  +  2c  -  Z^'  -  2^  -  c'  -V-  a  +  c  -  1  +  ^. 


4,74  DIVISION  BY  POLYNOMIALS. 

29.  12  {x  -  yf  -3x{y  -z)-  2y  {x  +  z)  -  20z  {y  +  3z) 

-r-6{x  +  2z)-3y, 

30.  {4.x'  -  9y')  {Sx'  -  27y')  -~  {2x  -  Zy)\ 

31.  12  +  82^'  +  106^'  -  70^'  -  112a'  -  38  -r-  3  -  5«  +  7a'. 

32.  a^  (1)  +  c)  +  y'  (^  -  c)  +  c"  {a  —  I)  -\-  ale  -^  a  -^l  -^  c. 

33.  a'  +  V-\-c'-Mhc-^a  +  l)-\-c. 

34.  x^  —  (a  +^)  i^""  +  (5^  +  (^P)  X  —  aq  -^  X  —  a,         \ 

35.  a*  -  13a'  +  36  -^  a'  +  5a  +  6. 

36.  x''  +  x'y""  -\-y^  -T-x"^  —  Q:y  +  ^/^ 

37.  3a^  -  8a'^'  +  3aV  +  61'  -  W&  -v-  a'  -  l\ 

Z'^.  y'  -  3^V  +  3z/V  -  2;'  -^  ,^'  -  3y'cc  +  3yx'  -  x\ 

39.  16aV—  7 ate  —  c'—  ^^a^lx  —  ^a'h'' -^^acx  -^  Sax  —  6ab  —  c. 

40.  x'  +  (a'  -  2b')  x'  -  {a'  -  h')  x'-a'-  2a'b'  -  a'b' 

-■x'  -a'  -  y"' 

41.  6  {x^-^y^)  +  {V^xy  -  4)  (x-^y)  -%{x'-\-  y')  -  16xy  -  120 

-^'*+^'  +  2.T(l+^)  +  2,y  +  6. 

42.  a'  -b'  ^a^-  U. 

43.  a'b'"^  +  2acZ?^  +  "  +  2a:z:Z>^  +  c'V''  +  2^;^^;'*  +  .t' 

-V-  aZ^'"  +  cZ>"  +  X. 

44.  (^"-^^)(^^-^^)-^^  +  /. 

45.  20a'^Z>^  -  ^OSa'b''  -  121a''¥'  +  132a'"^'  +  245a^^Z>^' 

-^  9a'^^  -  16a^'  +  lla^Z^'. 

46.  1  +  34  a;*^  -  20x'  +  20.t'  -  4a;'  +  12:?;'  -  31^;^ 

-f-  2a;  +  4a;'  -  3a:'  +  1. 

47.  2a;3^  —  6a;2^^"  +  Qx^y^""  —  2^/^'^  -v-  a:"  —  y"". 

48.  a  (a  -  1)  a;'  +  (a'  +  2a  -  2)  a:'  +  (3a'  -  a')  x  -  a' 

-i-  aa;'  —  2a;  —  a'. 

49.  a'^'  —  ^  (a'  +  Z>)  2/  +  ^^^  -^  ay  —  b. 

50.  (a  +  J)  (a  +  6')  —  (a  +  Z>)  (^  +  c)  -^-  a  —  d 

51.  a;'  +  (4a^  -  J)'^)  x  ^  {a  -  2b)  (a'  +  db')  ^  x  -  a  +  2h 

52.  a;' —  ^~^^  o:^  +  j^"^' 

53.  *  -  6^'  +  27^^  ^  i  +  2^  +  3z\ 

54.  a^'^-^^'W^c  —  a2'^  +  '^-^^^"^6'^  +  a-''^"^c"*  +  a^^-'*Z>^^  +  ^c'* 


FRACTIONS.  475 

Fractions. 

§  108. 

Execute  the  following  multiplications  of  fractions  by  entire 
quantities  by  dividing  the  denominators: 


X  a—h.  2.     3       7^  X  « +  ^. 


a-h'^  a'  -  ¥ 

m^  jf  -\-  Q^ 

a^  —  ¥  ,    ,  „      7)1"  -\-n'' 

a  -\-ab  m  —  m  n  ^  ' 

9.  z \ ^3 4  X  1  —  m'.   10.  — - — 3  x\-\-x. 

Execute  the  following  multiplications  by  dividing  the 
denominator  by  one  factor  of  the  multiplier,  when  denomi- 
nator and  multiplicator  have  a  common  divisor,  and  then 
multiplying  the  numerator  by  the  other  factor  of  the  multi- 
plier: 

Here  the  denominator  is  {a  —  by,  and  the  multiplier  is 
(a  —  b)  {a  +  b).  We  multiply  by  (a  —  b)  by  dividing  the 
denominator,  and  by  a-{-b  by  multiplying  the  numerator, 
m  (  «  4-  b) 


Hence  the  product  is 


a  —  b 

-5  X  m^  —  n^. 


3.  — ^ — -; —7-^  X  m^  —  4:n^. 

m'  —  4:7nn  -\-  4:n 

6.  — ^ — \-T-nr  X  «'  -  *'• 

ax  +  a?/  -\-  ox  -\-  by 


476  Fll ACTION  8. 

1 


Xa'  +  h\ 


*'  ax  —  ay  -\- Ix  —  ly 

1 

8. X  mx  +  mil  —  nx  —  nil. 

mx  —  nx  —  my  -\-ny  ^  ^ 

Execute  the  following  divisions  by  dividing  the  numerator 
by  as  many  factors  of  the  divisor  as  possible,  and  multiplying 
the  denominator  by  the  remaining  factors: 

ax  .  X  my 

~  ar.     Ans,  .  2.  -~-  -r-  mq, 

mn  mnr  Zpr         ^ 


3 
5 
7 
9 
II 


ah       ^  a"  -\-al)        ^      ^^ 

mn  a  —  0 

m  -{-  p  -^  X  ~  2xy  -\-  y 

cx-\-  cy         2         2  o    ^^  ~~  ^y       t  \2 

ax  —  ay  ^  ox-\-oy^        ^ ' 

— '--rr-  -^  («  +  ^hy.  10.  '—-  -^  a"  —  h\ 

a  —  zb  ^  '  a  —  0 

ac  —  he 


ax  -\-  Ix 


h\ 


Execute  the  following  indicated  multiplications  or  divi- 
sions, and  aggregate  each  product  or  quotient  into  a  single 
fraction: 

la    ,'b\la_h\  2    fl  __  i")  1. 

wi       nl  \m       nj'  '    \a       h )  c' 


3 
5 
7 
9 
II 


m    \      m 


-\-  X  \'        l  —  mJl-\-m 

\a-{-b       a  —  hi  a  ^  ^  \a  —  h         J 

\m       nJ  \m       nJ  \  ml 

f  1 4.  Z  _  V!}:\  f  1  _  Z  _  !?_^ 

\        m        pJ  \        m        pj' 


FRACTIONS, 


477 


13 
14, 

16 


21 


\    ^ ml  \7n  /   '       J  \^^'       my 

(m  +  n)  ( h  -)  —  (^^  —  ^0 • 

^      ^     ^  \m       nJ       ^  \m       nl 

(-^  +  -^  +  -^)  {x  +  y^z). 
\x  —  yy  —  zz  —  xj^ 


^-  771  +  ^. 

w       n 


17. 


m  —  71, 


a  +  l 


a  +  b 
c'  +  h:' 


a  —  b   '  a 


19 


a b       ,   a'  +  b' 

'  a+  b      a  —  b  '    a  —  b 


'^x'^  x'~^  x'   '  x'^  x'' 


Factor  th^  following  fractional  expressions: 


c  X 

'  x^       c^' 

^'  ¥  ~^  xy  ~^  y'' 

1  +  a' 

^'  2c'       Sx'' 
9.  S--3I  +  2. 


h' 


h 


2. 

a'       V 

4. 

x'      /■ 

/; 

m  ip'  +  q') 

n{f-^n^<t)' 

8. 

m       ,  m® 

4-3-. 

71         n^ 

10, 

4  +  5<  +  4^. 

110. 


Reduce  the  following  complex  fractions  to  simple  ones: 


a^  b 


478  SIMPLE  EQUATIONS. 


c     .    d       X 

15+-  +  ~  1  +  m 


d       X       c 


1_ 

abc 


r  —  s 


r  -\-  s      r  —  s 


1' 


1.1,1  1  + 
c        d       X 

a  a 

T  e.  1. 

C  0 


m 


1.1.1 


1,1,1'  _  +  _  +  . 

— r  7-  +  -  m       n      p 
a       0        c  ^ 

r  -\-  s  r 


1+-^ 


.+' 


r  -{-  s      r  —  s  r 

r  -\-  s 


1 

m 

+i 

tl 

1 

^h 

H- 

mnp 

1 

1 

m 

n 

b        ^       a  14.      2         3. 

1  --  -r-  m        n 

Equations  of  the  First  Degree  with  One 

Unknown  Quantity,  x. 

§  139. 

n  m 

(IX  ox  ex  2       1       72       I  9 

DC       ac      ah  '         ' 

3.   {^a  —  x)  {h-\-x)-=zx{h  —  x). 

hx    .     ax  a    ,    xa 

5-  C5'a;---  +  ca;  =  --  +  (a  +  c)a!-c. 


^'^  =  ^+  a   '   de 


ONE   UNKNOWN  QUANTITY.  473 

he        cfx 


=  1. 


'  a  —  h      a  -\-h 

_^      GX  CX  U/X  _ 

'•  7  +  T  +  T-^  =  ^'- 

9.  —  ao  +  —  ac  —  —  ca;  =  —  ac  +  2ao  —  bcx. 
0  o  o  4 


II. 

ah  _ 

hc  +  d+-, 

X 

12. 

3ri  + 

X 

X 

,  m(a  —  x) 

1  X  4-  a 

14,  - 


15. 


X  -\-  a       X  —  a       a^  —  x^' 

.7;  —  1  ~~       r?;  +  1  x'^  —  1 


16. —  • -\ =  0. 

X  —  4:       X  —  Q       X  —  "Z 

dahc  aW  {2a  +  b)  Fx  _  ^ 

'^*  a-\-h'^  {a  +  hf^   u{a  +  by    -^^^+  a' 

18.   (a  +  a:)  (Z^  +  :r)  -  6^  (Z^  +  c)  =r  ^  +  a;\ 

19. 7 h- h^  =  0. 

6?  6>  c 

X  -\-  a       X  4-  a^   ,   X  -\-  a^       x  -\-  a^ 

a  a^       '       a^  a^ 

21.  [ax  +  ^)  (Z>.^'  —  a)  —  {ax  —  Z>)  (Z^.-r  +  «)  =  (i^  -f  Z>, 

«  +  a;  Z^  +  a; 

22.  ^''^ = — ■  —  ah . 

0  a 

1  ,7)1  —  71  1  ,    771  -\-  n 

23. = ■ — . 

m  —  71  x  711 -{-n  X 

1  p 

24.  - 


{771  -\-  Olf         771-^71         2  {7)1  +  71)' 

o 1 

p  X  p 


480 

25.  I  = 

26.  m  — 
27 


SIMPLE  EQUATIONS. 


nx 


1  --  7nx 

P_±J^^ 

q  +  x       q-\-x' 
1 

:  + 


m. 


ah  —  ax       he  —  hx       ac  —  ax' 
28.  (m  +  nV  =  3m'  +  ?i'  —  -'^ — . 

X 


30.   ^^  = 


^^~g^       ^6?(^  +  c) 


771  m 


1  +  :^       l-x     ■ 
31. =  1. 


n 


n 


m-\ — 

X 

32. =  2/w. 


1  +  ic       1  —  a; 
1  +  ^-^ 


ZZ- 


X  -\-  a 


m 


1  + 


=  ^. 


34. 


X  -\-  a 


1  + 


=  2^. 


Equations  of  the  First  Degree  witli  Two 
Unknown  Quantities. 

§§137-140. 


j  mx  —  ny  —  0. 


^  +   y  =  a. 

I  fl^o;  +  a^y  =  ap, 
[  hx  -j-  h'^y  =  ^^. 


\px  +  qy  =  a. 
\  X  —   y=-h. 


4.   < 


y 


h  -{-  y         3a -\-  x' 
ax  +  '^hy  —  d. 


1 

x-\-y^3' 

-^=3. 

y 


^     \hx^  y-a. 
\    X  -\-hy  •=  2a. 


TWO    UNKNOWN  QUANTITIES. 


481 


X         _  y 

x-\-y:=s. 

a    ,   h 

-  +  -  =  c. 
X       y 

--{--  z=zd. 

^x       y 


12. 


a        h  __ 
X       y 

\,x       y 
b  {x  —  a)  -\-  a  {y  —  h) 

X  —  a  :  y  —  b  =  b  :  a. 


13.  iix-{-b  =  my  -{-  d  =  c.         14.  mx  =  ny  —p  =  x  -\-  qy. 
15.  m{x  +  y)  =  71  {x  —  y)  =  r. 
X  ^      _      1 


16. 


la  +  b      a  —  b 


a  —  b       a-\-b' 

y    _    1 
6' 


*7-  1 


x-\-y 
1 


+ 


1 


1-x- 

1 


L 1  —  ic  +  .V 


2 

3* 

4 

3* 


Equations    with    Three    or   More    Unknown 
Quantities. 


[x-{-y-^z  =  a. 
mx  =  ny. 

[py  =qz. 

(  X  -\-  y  =  a. 

}y-\-z  =  2a. 
[z~i-x  =  da-\-b. 

i  ¥  =  iy- 

iiz=ix  +  l. 

(x  —  z  =  am. 
)y-^z  =  bm. 
(^x  —  y=  cm. 

(x  =  t/  —  2z. 
\y  =  '3z-  2x. 
(z=y  +  l. 


6. 


x+y+z=i 
X  _y  _z 
a       b       c' 


m 


n 


-  -  -  _  ^. 

X        y       z' 

^  +  y  +  ^  =  S' 

a;  +  ^  +  ^  ==  30. 
8a;  +  iy  +  2z  =  50. 
272;  +  '9y  +  3z=  64. 

''x-\-y  -\-  z  =  au. 
x-}-y  =  bu. 

X 

—  =  m. 

7ix  +  y  -{-  z  =  a. 

X  -|-  ny  -j-  z  =^  b. 
X  -{-  y  -{-  nz  =  c. 


482 


8IMFLE  EQUATIONS. 


II.    < 


IS- 


IS- 


f  2;  +  2?/  =  8. 
^  +  2^  ==  12. 
z  -{-u    =    8. 


(  «^  +  ^^  +  cz  —  d. 

\a^x-\-¥y-^ez=d\       14. 


17.    i 


19.    i 


2a;  +  3^/  +  52;  =:  67. 

2a;  +  3^  4- 4^  =  35.    16. 

2^  -  3«/  +  5^  =  13. 


a;  4-  ^  =  fl^. 

y^z^l,  ] 

z^  —  X  =^  d, 

-hx  -\-ay  _        a  —  l 

{b—c){a—cy 
h  —  c 


cy 

c 
+  hz 

az 

a 
+  ex 

{c--a){b—ay 

c  —  a 
[a-b){c-b)' 


a  a  —  r  a  —  s 
^  "^  ^  -  r  "^  Z^  -  ;^ 
c       c  —  r      c  —  Si 

X       y        a 

.-  +  -  =  -• 

^y       z       c 

3^  +    2/  +    ^  =  3. 

X  -\-  4:y  -\-    z  =  4t. 
X  -\-   y  +  6z  =  5, 


=  1. 
=  1. 


b+  c 

y 


+ 

+  ■ 


y 


c  ~  a 

z 


a  +  b. 


a  -\-  a   '    a  —  b 

z  X 


a  +  b 


:=b  +  C, 

—  c  -\-  a. 


PROBLEMS     LEADING    TO     EQUATIONS     WITH     ONE 
UNKNOWN     QUANTITY.* 

1.  A  capitalist  earned  4  per  cent  interest  fi*om  f  of  his  in- 
vestment, and  5  per  cent  from  the  remaining  \,  making  a 
total  annual  interest  of  $2940.    What  was  tlie  amount  invested? 

2,  What  quantities  must  be  added  to  each  term  of  the 

TYl 

fraction  —  that  it  may  take  the  following  series  of  values: 


*  Although  only  one  unknown  quantity  is  really  necessary  in  these 
problems,  the  student  may  often  find  it  convenient  to  use  two  or  more. 


niOBLEMS.  483 

AVhat  qiifintities  musfc  be  subtracted  from  each  term  to  pro- 
duce tlie  same  results  ?  Explain  the  relation  between  the 
answers  in  the  two  cases. 

3.  A  man  is  40  years  old,  and  his  wife  is  36.  In  how  many- 
years  will  the  sum  of  their  ages  be  5?  Explain  the  results 
when  Ave  put,  in  succession, 

s  =  100;         s  =  76;         and         s  =  50. 

4.  A  railway  train  passed  a  station  at" the  speed  of  m  miles 
an  hour.  Then  k  hours  later  another  passed  in  the  same 
direction,  going  n  miles  an  hour.  Supposing  the  speeds 
uniform,  at  what  distance  and  at  what  time  did  they  meet? 
Explain  the  relation  of  the  answers  when  m  >  n  and  when 

7)1   <   71, 

5.  If,  in  the  preceding  problem,  the  second  train  went  in 
the  opposite  direction,  what  would  the  answer  be?  Explain 
the  relation  between  the  answers. 

6.  A  ship  sailed  from  port  with  a  speed  Ic  knots  per  houi*. 
In  7i  hours  after  sailing  she  was  followed  by  a  steamer,  who 
overtook  her  in  7i  hours.     What  was  the  speed  of  the  steamer? 

7.  An  oarsman  who  pulls  6  miles  an  hour  rows  from  his 
house  down  a  river  whose  current  is  2  mi.les  an  hour,  and  re- 
turning gets  back  3  hours  after  he  started.  How  far  did  he 
go? 

8.  On  the  same  stream  one  rower  pulling  6  miles  an  hour 
going  down  stream,  and  another  pulling  7  miles  an  hour 
going  up  stream,  started  out  at  the  same  moment;  but  the 
starting-point  of  the  second  was  5  miles  below  that  of  the 
first.     At  what  point  and  in  what  time  did  they  meet? 

9.  A  steamer  goes  down  the  Rliine  from  Mayence  to  Co- 
logne, 117  miles,  in  8^  hours,  but  requires  14  hours  f6r  the 
return  journey.     What  is  the  speed  of  the  current? 

10.  On  an  ocean  the  crests  of  the  waves  are  y^  of  a  mile 
apart,  and  are  moving  at  the  rate  of  40  miles  an  hour.  If  a 
ship  steams  15  miles  an  hour,  how  many  times  an  hour  will 
she  pitch  when  going  with  the  waves,  and  how  many  times 
when  goinsr  aarainst  them? 


484  SIMPLE  EQUATIONS. 

11.  A  number  is  divided  into  three  parts,  of  which  one  is 
30  less  than  a  half,  a  second  10  less  than  a  third,  and  the  re- 
maining part  8  greater  than  a  fourth.  Find  the  number  and 
the  three  parts. 

12.  From  a  line  was  taken  J  its  length  and  2  feet  more, 
and  from  what  was  left  ^  its  length  and  2  feet  more,  leaving 
^  the  whole  line  and  2  feet  more.  What  was  the  length  of 
the  line? 

13.  A  team  performed  a  journey  in  8  hours,  going  one 
third  the  way  at  the  rate  of  25  miles  an  hour,  and  the  remain- 
ing two  thirds  at  the  rate  of  40  miles  an  hour.  What  was  the 
disitance? 

14.  A  grocer  has  60  pounds  of  tea  worth  75  cents  a  pound, 
formed  by  mixing  one  kind  worth  80  cents  a  pound  with 
another  worth  50  cents  a  pound.  How  many  pounds  of  each 
kind  were  in  the  mixture? 

15.  Divide  a  line  of  length  /  so  that  |  of  one  part  shall  be 
equal  to  f  of  the  other  part. 

16.  A  man  is  6  years  older  than  his  wife.  Ten  years 
hence  the  sum  of  their  ages  will  be  7  times  the  age  of  the  wife 
14  years  ago.     What  are  their  ages? 

17.  A  man  who  must  be  back  in  1  hour  starts  in  a  coach 
going  m  miles  an  hour,  and  walks  back  at  the  rate  of  n  miles 
an  hour.     How  far  can  he  go  and  be  back  in  time? 

18.  The  earth  performs  a  revolution  round  the  sun  in  1 
year;  Mars,  in  1|  years.  What  is  the  mean  interval  between 
conjunctions;  ihafc  is,  between  the  times  at  which  the  earth 
passes  Mars? 

19.  The  periodic  time  of  Jupiter  is  11-f  years;  of  Saturn, 
29|-  years.  At  what  intervals  will  the  earth  be  in  conjunction 
with  each  of  them,  and  at  what  intervals  will  they  be  in  con- 
junction with  each  other? 

20.  Two  persons,  A  and  B,  were  mounting  a  tower,  B  be- 
ing always  24  steps  behind  A.  When  A  was  half  way  up  he 
said  to  B,  "  When  I  reach  the  top,  you  will  be  8  times  as  high 
as  you  are  now."     What  was  tlie  height  of  the  tower? 

21.  The  circumference  of  the  front  wheels  of  a  carriage  is 
9  feet;  of  the  hind  wheels,  12  feet.     How  far  has  the  carriage 


PBOBLEMS.  485 

driven  when  the  front  wheels  have  made  m  turns  more  than 
the  hind  ^wheels? 

2  2.  The  members  of  a  club  have  to  raise  a  certain  sum  of 
money.  If  each  member  contributes  $2,  there  will  be  $28  too 
much;  if  $1.25,  there  will  be  $32  too  little.  How  many  mem- 
bers are  there,  and  what  is  the  amount  to  be  raised? 

23.  If  a  dealer  sells  a  piece  of  cloth  at  7n  cents  a  yard,  he 
gains  d  dollars;  if  at  7i  cents  a  yard,  he  loses  c  dollars.  What 
is  the  length  of  the  piece,  and  the  purchasing  price  per 
yard? 

24.  A  merchant  by  the  profits  of  trade  increases  his  capi- 
tal each  year  by  20  per  cent  of  the  amount  at  the  beginning, 
but  takes  out  $1000  at  tlie  end  of  each  year  for  his  board.  At 
the  end  of  the  third  year  he  has  increased  his  capital  by  $200 
more  than  -|  of  its  original  amount.  With  what  amount  did 
be  start? 

25.  A  boat  which  steams  12  miles  an  honr  makes  her  (rip 
in  3  kours  going  down  stream,  and  in  5  houi-s  going  up  stream. 
W^hat  is  the  speed  of  the  current  and  the  length  of  the  trip? 

26.  A  number  is  increased  by  n,  and  the  sum  multiplied 
hyn\  this  product  is  then  increased  by  71,  and  the  sum  multi- 
plied by  n,  with  the  result  2?^.^     What  is  the  numbei'? 

27.  A  number  is  diminished  by  n,  and  the  remainder  multi- 
plied by  n  ;  the  same  operation  is  repeated  on  the  product, 
and  again  repeated  on  the  second  product,  with  the  result 
—  w^     What  is  the  number? 

28.  What  number  is  that  whose  fourth  part  exceeds  its 
sixth  part  by  2? 

29.  If  you  add  4  to  a  certain  number,  the  sum  is  2  less 
than  twice  the  number.     What  is  it? 

30.  Divide  $520  among  three  people  so  that  the  first  may 
have  $20  less  than  the  second,  and  tlie  second  $10  more  than 
one  fourth  the  share  of  the  third.     What  must  each  receive? 

31.  Divide  c  dollars  among  three  people  so  that  the  first 
may  have  a  dollars  less  than  the  second,  and  the  second  m 
dollars  more  than  one  fourth  the  share  of  the  third.  What 
must  each  receive? 

32.  A  left  a  certain  town   at  G  miles  an  hour,  and  in  8 


486  SIMPLE  EQUATIONS. 

hours  after  was  followed  by  C  at  8  miles  per  hour.     In  how 
many  hours  did  0  overtake  him? 

33.  A  left  a  certain  town  at  b  miles  an  hour,  aud  in  n 
hours  after  was  followed  by  D  at  c  miles  per  hour.  In  how 
many  hours  did  D  overtake  him? 

34.  A  farmer  said,  if  he  had  5  more  sheep,  and  sold  them 
at  $4  each,  he  would  have  5  times  as  many  dollars  as  he  now 
has  sheep.     How  many  sheep  has  he? 

35.  A  farmer  said,  if  he  had  a  more  sheep,  and  sold  them 
all  at  n  dollars  each,  he  would  have  c  times  as  many  dollars  as 
he  now  has  sheep.     How  many  sheep  has  he? 

36.  If  you  divide  my  age  10  years  hence  by  my  age  10  years 
ago,  you  will  get  the  same  quotient  as  if  you  should  divide 
my  present  age  by  my  age  15  years  ago.  What  is  my  present 
age? 

37.  If  you  divide  my  age  c  years  hence  by  my  age  a  years 
ago,  you  will  get  the  same  quotient  as  if  you  should  divide 
my  present  age  by  my  age  d  years  ago.  What  is  my  present 
age? 

2,^,  Divide  $415  among  A,  B  and  C  so  that  A  shall  have 
$40  less  than  B,  and  C  $20  more  than  half  as  much  as  A  and 
B  together. 

39.  Divide  %a  among  C,  D  and  E  so  that  C  shall  have  $m 
less  than  D,  and  E  %7i  more  than  one  third  the  share  of  0  and 
D  together. 

40.  A  can  do  a  piece  of  work  in  20  days,  B  in  24  days,  and 
C  in  30  days.     In  what  time  can  they  together  do  the  work? 

41.  A,  B  and  0  can  do  a  piece  of  work  in  4  days,  A  alone 
in  12  days,  and  B  alone  in  10  days.  How  long  would  it  take 
C  to  do  it? 

42.  A,  B  and  0  can  do  a  piece  of  work  in  6  days,  A  alone 
in  9  days,  and  B  alone  in  12  days.  How  long  would  it  take  C 
to  do  it? 

43.  A  can  do  a  piece  of  work  in  a  days,  B  in  Z>  days,  and 
C  in  c  days.     In  what  time  can  they  together  do  it? 

44.  A  man  is  12  years  older  than  his  wife;  four  years  ago 
8  times  her  age  was  5  times  his.     What  arc  their  present 


PROBLEMS.  487 

45.  A  man  is  a  years  older  than  his  wife;  b  years  ago  c 
times  her  age  was  m  times  liis.  What  are  their  present 
ages? 

46.  Divide  $1200  profit  so  that  A  may  have  one  fourth 
and  $100  more,  B  $5*0  less  than  one  third,  and  0  $'^50  moro 
than  one  sixth. 

47.  The  interest  on  -^^  of  a  certain  capital  at  5  i)er  cent 
added  to  the  interest  on  the  remainder  at  6  per  cent  is  equal 
to  $1680.     What  is  the  capital? 

48.  A  person,  asking  the  distance  to  a  certain  city,  was 
told  that  after  he  had  gone  one  fourth  the  distance  and  two 
thirds  the  remaining  distance,  he  would  still  have  20  miles  to 
travel.     What  was  the  distance? 

49.  How  far  can  a  person  who  has  5  hours  to  spare  ride 
at  6  miles  per  hour  so  as  to  walk  back  in  time  at  4  miles  per 
hour? 

50.  How  far  can  a  person  who  has  n  hours  to  spare  ride 
at  1)  miles  per  hour  so  as  to  walk  back  in  time  at  c  miles  per 
hour? 

51.  A  man  bought  15  horses  for  $1665,  paying  $120  for 
each  good  horse,  and  $75  each  for  the  poor  ones.  How  many 
of  each  did  he  buy? 

52.  The  difference  of  the  squares  of  two  consecutive  num- 
bers is  15.     What  are  the  numbers? 

53.  The  difference  of  two  numbers  is  2,  and  the  difference 
of  their  squares  is  28.     What  are  the  numbers? 

54.  The  sum  of  two  numbers  is  12  ;  the  square  of  the 
greater  is  48  more  than  the  square  of  the  less.  What  are  the 
numbers? 

55.  The  product  of  two  consecutive  numbers  is  4  more 
than  the  square  of  the  less.     What  are  the  numbers? 

56.  Divide  60  into  three  such  parts  that  one  third  of  the 
first,  one  fourth  of  the  second,  and  one  fifth  of  the  third  shall 
be  equal  to  each  other? 

57.  Divide  80  into  four  such  parts  that  if  the  first  be  in- 
creased by  3,  the  second  diminished  by  3,  the  third  multiplied 
by  3,  the  results  shall  be  equal. 

58.  The  greater  of  two  numbers  is  4  times  the  less;  if  each 


488  SIMPLE  EQUATIONS. 

be  increased  by  3,  the  greater  will  be  3  times  the  less.     What 
are  the  numbers? 

59.  A  man  is  10  years  older  than  his  wife;  in  10  years 
twice  the  sum  of  their  ages  will  be  6  times  her  present  age. 
What  is  the  age  of  each? 

60.  A  man  bought  a  certain  number  of  sheep  for  $1200; 
he  reserved  80,  and  sold  the  remainder  for  $960.  How  many 
did  he  buy? 

61.  A  father  aged  48  years  has  a  son  aged  12.  In  how 
many  years  will  the  age  of  the  father  be  three  times  that  of 
the  son? 

62.  A  merchant  has  two  kinds  of  tea;  one  cost  $1.50  a 
pound,  and  the  other  $2.  He  wishes  to  mix  them  so  as  to 
have  50  pounds  worth  $1.80  a  pound.  How  much  of  each 
must  he  use? 

67,,  In  a  certain  quantity  of  mortar  the  sand  was  15  pounds 
more  than  f  of  the  whole,  the  lime  9  pounds  less  than  ^  of  the 
whole,  and  the  plaster- of -paris  6  pounds  less  than  \  the  sand. 
What  was  the  amount  of  the  mortar? 

64.  A  laborer  agreed  to  work  50  days  on  the  condition 
that  he  should  receive  $1.50  for  everyday  he  worked,  and  for- 
feit $0.75  for  every  day  he  was  idle.  At  the  end  of  the  time 
he  received  $48.     How  many  days  did  he  work? 

65.  A  grocer  having  60  pounds  of  coffee  worth  15  cents  a 
pound  mixed  it  with  so  much  coffee  at  18  cents  a  pound 
that  the  mixture  was  worth  16  cents.     How  much  did  he  use? 

66.  The  interest  on  a  certain  capital  at  5  per  cent  is  $20 
less  than  the  interest  on  $900  more  at  1  joer  cent  less.  What 
is  the  capital? 

67.  A  woman  bought  200  apples  at  5  for  3  cents,  and  sold 
part  at  2  for  a  cent,  and  part  at  5  for  4  cents,  thereby  making 
10  cents.     How  many  of  each  kind  did  she  buy? 

68.  A  and  B  play  at  cards.  A  begins  with  $120,  and  13 
with  $180  ;  when  they  stop  playing  B  has  four  times  as  much 
as  A.     How  much  did  B  win? 

69.  From  a  cask  of  wine  one  fourth  leaked  out,  then  20 
gallons  were  drawn,  when  it  was  found  to  be  10  gallons  less 
than  half  full.     How  much  did  it  hold? 


PROBLEMS.  489 

70.  An  estate  of  $4680  is  to  be  divided  among  4  sons  and 
3  daughters.  Each  son  is  to  receive  $40  more  than  the  next 
younger;  the  eldest  daughter  is  to  have  $20  less  than  the 
eldest  son.  and  each  of  her  sisters  $20  less  than  the  next 
older.     What  did  each  child  get? 

71.  A  sum  of  $2880  is  to  be  divided  among  A,  B  and  C. 
Five  times  A's  share  is  to  be  equal  to  three  times  C's,  and  B  is 
to  have  twice  as  much  as  A  and  C.     What  does  each  receive? 

72.  Six  plasterers,  8  journeymen  and  12  apprentices  re- 
ceive at  the  end  of  a  certain  time  $387.50.  The  plasterers 
receive  $2  a  day,  the  journeymen  $1.25,  and  the  apprentices 
75  cents.     How  many  days  did  they  work? 

73.  In  the  above  problem,  what  should  each  class  of  work- 
men receive  if  each  plasterer  worked  3  days  more  than  the 
journeymen,  and  the  apprentices  6  days  less? 

74.  A  man  wished  to  give  10  cents  each  to  some  beggars, 
but  found  he  had  not  enough  of  money  by  14  cents;  he  then 
gave  each  one  8  cents,  and  found  that  he  had  10  cents  re- 
maining.    How  many  beggars  were  there? 

75.  A  post  is  6  feet  more  than  J  in  the  mud,  2  feet  less 
than  ^  in  the  water,  and  4  feet  in  the  air.  What  is  the  length 
of  the  pole? 

76."  A  and  B  begin  trade.  A  has  $1000,  and  B  $1210. 
The  former  gains  a  certain  per  cent  on  his  investment,  and 
the  latter  loses  the  same  per  cent,  when  their  capitals  are  found 
to  be  equal.     What  was  the  amount  lost  and  gained? 

77.  A  person  in  play  lost  ^  of  his  money,  then  won  $60, 
after  which  he  lost  \  of  what  he  then  had,  when  he  found  he 
had  but  $350  remaining.     Wliat  had  he  at  first? 

78.  In  a  camp  of  3294  soldiers  there  were  3  cavalry  to 
every  26  infantry,  and  half  as  many  artillery  as  cavalry. 
What  was  the  number  of  each? 

79.  The  right-hand  digit  of  a  certain  number  is  2  less  than 
the  second;  and  if  the  number  be  divided  by  the  sum  of  the 
digits,  the  quotient  will  be  7.     What  is  the  number? 

80.  The  length  of  a  town  lot  exceeds  its  width  by  12  feet. 
If  each  were  3  feet  greater,  there  would  be  an  increase  of  645 
square  feet  in  its  dimensions.     What  is  the  length? 


490  •  SIMPLE  EQUATIONS. 

8 1.  A  house  was  sold  for  $0800,  by  which  there  was  a  cer- 
tain gain.  If  it  had  been  sold  for  $1000  less,  3  times  the 
resulting  loss  would  liave  been  twice  the  present  gain.  What 
was  the  cost  of  the  house? 

82.  A  can  do  a  piece  of  work  in  12  days,  and  B  in  15. 
After  A  has  worked  4  days  B  comes  to  help  him.  In  what 
time  can  they  both  finish  it? 

St,,  a  tank  has  two  filling  and  one  emptying  pipe.  One 
can  fill  it  in  12  hours,  the  other  in  24  hours  ;  and  the  third 
can  empty  it  in  18  hours.  If  tliey  are  started  at  the  same 
time,  how  long  will  it  take  to  fill  the  tank? 

84.  In  the  preceding  problem,  suppose  the  third  can 
empty  it  in  8  hours.     How  long  will  it  take  to  fill  it? 

85.  Suppose  it  is  full  already,  and  the  third  can  empty  it 
in  6  hours.     How  long  will  it  take  to  empty  it? 

S6,  A  person  travelled  168  miles,  of  which  he  went  3  by 
boat  and  4  by  coach  to  every  6  by  rail,  and  walked  one  third 
as  far  as  he  went  by  boat.  How  many  miles  did  he  travel  by 
each? 

87.  The  sum  of  two  numbers  is  42.  If  the  less  be  divided 
by  the  greater,  the  quotient  will  be  less  by  ^  than  when  the 
less  is  divided  by  half  the  greater.  What  are  the  num- 
bers? 

2>S,  A  and  B  are  of  the  same  age.  Three  times  A's  age 
6  years  ago  is  equal  to  twice  B's  age  9  years  hence.  What  is 
the  age? 

89.  In  tossing  pennies,  A  threw  heads  3  times  out  of  5, 
and  B  4  times  out  of  7.  In  all  they  get  41  heads.  How  many 
times  did  they  toss? 

90.  What  two  numbers  are  those  whose  sum  is  13,  and 
whose  product  added  to  the  square  of  the  less  makes  50? 

91.  A  tank  has  five  pipes.  No.  1  can  fill  it  in  6  hours,  No. 
2  in  8  hours,  and  No.  3  in  12.  No.  4  can  empty  it  in  9  hours, 
and  No.  5  in  18.  If  they  begin  at  the  same  time,  how  long 
will  it  take  to  fill  the  tank? 

92.  A  starts  from  a  certain  place,  and  travels  at  the  rate 
of  17  miles  in  5  hours.     One  hour  and  53  minutes  after,  B 


PROBLEMS.  491 

starts  at  the  rate  of  19  miles  in  4  hours.     How  far  will  thoj 
travel  before  B  overtakes  A? 

93.  Two  persons  start  from  the  same  place  at  the  same 
time^  going  in  the  same  direction.  One  travels  2  miles 
an  hour  faster  than  the  other.  After  they  had  gone  as  many 
hours  as  tlie  slower  goes  miles  per  hour,  their  distance  apart 
was  equal  to  half  the  distance  travelled  by  the  faster.  How 
long  did  they  travel? 

94.  Two  men  travel  in  opposite  directions;  the  rate  of 
one  is  1  mile  ntore  than  two  thirds  the  rate  of  the  other. 
When  they  had  gone  4  hours  the  distance  apart  was  equal  to 
44  miles.     What  were  their  rates? 

95.  An  officer  in  arranging  his  men  in  the  form  of  a  square 
found  that  he  needed  5  men  to  complete  the  square,  and  by 
increasing  the  file  by  6  and  diminishing  the  rank  by  5  he  had 
5  men  too  many.     How  many  men  had  he? 

96.  A  coach  that  travels  6  miles  an  hour  starts  50  minutes 
after  another  that  goes  5  miles  an  hour.  How  far  will  the 
first-named  travel  in  order  to  be  11  miles  ahead  of  the  other? 

97.  A  merchant  withdrew  from  his  capital  $500  at  the  end 
of  each  year  for  current  expenses;  his  profits  each  year  were 
33^  per  cent  of  his  unexpended  capital.  In  3  years  his 
original  stock  was  doubled.     What  was  his  original  stock? 

98.  What  fraction  is  that  whose  denominator  is  2  more 
than  the  numerator,  and  if  3  be  subtracted  from  both  numer- 
ator and  denominator  the  fraction  will  be  f  ? 

99.  Divide  40  into  two  such  parts  that  the  greater  dimin- 
ished by  4  and  divided  by  the  less  increased  by  6  shall 
be  1|  ? 

100.  On  a  note  interest  is  paid  at  6  per  cent.  At  the  end 
of  the  first  year  $200  is  credited  on  the  principal,  and  the  rate 
of  interest  is  reduced  to  5  per  cent,  when  the  annual  interest 
is  diminished  by  one  fifth.     What  was  the  face  of  the  note? 

loi.  The  difference  between  the  simple  and  compound  in- 
interest  of  a  certain  principal  during  the  second  year  at  5  per 
cent  is  $10.     What  is  the  principal? 

102.  The  fore  and  hind  wheels  of  a  carriage  have  circum- 
ferences of  12  and  16  feet.      How  far  will  the  carriage  have 


492  SIMPLE  EQUATIONS. 

gone  when  the  sum  of  the  revolutions  made  by  the  wheels 
is  287? 

103.  During  the  first  3'ear  a  broker  gains  20  per  cent  on 
his  capital,  the  second  year  he  gains  30  per  cent  on  his  in- 
creased capital,  and  the  third  25  per  cent  on  his  re-increased 
capital,  when  he  finds  that  his  capital  is  $4910  more  than 
what  he  began  with.     What  was  his  first  capital? 

104.  A  man  sold  a  house  and  furniture  for  $6400;  f  of  the 
price  of  the  house  was  $200  less  than  f  the  price  of  the  furni- 
ture.    What  was  the  value  of  each? 

105.  A  purse  contains  65  coins,  i)art  cents  and  part  dimes. 
How  many  of  each  are  there  if  the  total  value  is  $2? 

106.  Each  member  of  a  base-ball  club  subscribes  as  many 
cents  as  there  are  members.  If  there  had  been  10  more 
members,  each  subscription  would  have  been  9  cents  less. 
How  many  members  were  there? 

107.  A  man  purchased  a  number  of  lemons  at  2  cents 
each,  and  f  as  many  at  3  cents  each;  he  sold  them  all  at  the 
rate  of  2  for  5  cents,  and  gained  25  cents.  How  many  of 
each  kind  did  he  purchase? 

108.  A  boy  in  flying  his  kite  lost  |  of  his  string,  then 
added  65  feet,  and  found  that  it  was  just  f  of  its  original 
length.     What  was  the  length  at  first? 

109.  A  and  B  start  from  two  towns  that  are  133  miles 
apart  and  travel  towards  each  other.  They  meet  at  the  end 
of  10  hours,  and  find  that  A  has  travelled  1^  miles  an  hour 
more  than  B.     How  many  miles  had  each  travelled? 

no.  A  man  owning  a  cow  and  horse  found  that  4  loads  of 
hay  would  keep  them  both  6  months.  Having  disposed  of 
his  horse,  ho  found  that  the  same  quantity  of  h-ay  would  last 
the  cow  14  months.     How  long  would  1  load  last  each? 

111.  A  has  $647,  which  is  $33  loss  than  4  times  what  B 
has;  0  is  worth  twice  as  much  as  A  and  B  together,  lacking 
$72.     How  much  have  B  and  C? 

112.  A  boat  which  could  move  14  miles  in  still  water  was 
accelerated  2^  miles  per  hour  going  down  stream,  and  retarded 
the  same  returning;  it  was  10  hours  longer  coming  up  a  cer- 
tain distance  than  going  down.     What  was  the  distance? 


PROBLEMS.  493 

113.  A  and  B  have  i\\%  same  income.  A  spends  |-  of  liis, 
and  B  by  spending  $200  a  year  more  than  A  finds  himself  at 
the  end  of  5  years  $450  in  debt.     What  was  their  income? 

114.  A  farmer  bought  22  cows  at  a  certain  price;  had  he 
paid  8  per  cent  less  he  could  have  purchased  1  more  cow  and 
had  $21  left.     What  was  the  price  of  each  cow? 

115.  A  son  is  -|-  the  age  of  his  father,  and  11  years  ago  lie 
was  f  of  his  age.     How  old  is  each  ? 

116.  A  man  rows  5  miles  an  hour  in  still  water.  How  far 
can  he  row  up  a  stream  and  back  in  3  hours,  the  stream 
flowing  a  mile  an  hour? 

1 17.  A  man  bought  some  sheep  for  $94.  Having  lost  7  of 
them,  he  sold  J  of  the  remainder  at  first  cost  for»$20.  How 
many  did  he  buy? 

118.  The  pei-imeter  of  a  rectangle  is  28  feet;  if  2  feet  be 
taken  from  its  length  and  added  to  its  breadth,  its  area  is  in- 
creased by  12  square  feet.     Find  its  original  breadth? 

119.  A  man  can  row  9  miles  an  hour  with  the  stream, 
and  3  against  it.  How  far  can  he  go  so  as  to  be  back  in 
6  hours? 

1 20.  The  first  digit  of  a  certain  number  exceeds  the  second 
by  5,  and  if  the  digits  be*  inverted  the  new  number  will  be  f 
of  the  original  number.     What  is  the  number? 

121.  Divide  $900  in  two  such  parts  that  the  interest  on 
one  part  at  4|-  per  cent  may  exceed  that  on  the  other  at  3|^  per 
cent  by  50  cents. 

122.  How  much  foreign  brandy  at  $8  a  gallon  and  whisky 
at  $3  a  gallon  must  be  mixed  together  so  that  the  compound 
may  be  sold  for  $9,  and  the  merchant  thereby  gain  30  per  cent. 

123.  A  person  has  two  kinds  of  coins.  Four  pieces  of  one 
make  a  dollar,  or  10  pieces  of  the  other.  How  many  of  each 
must  be  taken  so  as  to  have  7  pieces  equal  a  dollar? 

124.  Find  two  numbers  whose  product  is  72,  and  whose 
difference  multiplied  by  the  greater  is  found  by  subtracting 
the  product  from  18  times  the  greater. 

125.  A  person  after  spending  $200  more  than  ^  of  his  in- 
come had  remaining  $75  less  than  ^  of  it.  What  was  his  in- 
come? 


494  •  SIMPLE  EQUATIONS, 

126.  Divide  77  into  two  such  parts  that  the  quotient  of 
the  first  divided  by  8  added  to  tlie  quotient  of  the  second 
divided  by  9  shall  be  9? 

127.  The  sum  of  three  numbers  is  155.  If  the  second  be 
divided  by  the  first,  the  quotient  is  2,  and  2  for  a  remainder, 
and  the  third  divided  by  the  second  gives  3  for  a  quotient  and 

3  for  a  remainder.     What  are  the  numbers? 

128.  At  a  ball  there  were  twice  as  many  gentlemen  as 
ladies.  When  8  couples  danced  there  were  remaining  three 
times  as  many  gentlemen  as  ladies.  What  was  the  number  of 
each? 

129.  A  can  build  7  cubic  yards  of  wall  in  4  days,  B  12 
yards  in  5  days,  and  C  9  yards  in  2  days.  How  long  will  it 
take  all  three  to  build  850  yards? 

130.  Each  of  the  three  digits  of  a  certain  number  is  greater 
than  the  next  folloAving  by  1;  when  the  digits  are  inverted, 
the  new  number  will  be  18  more  than  ^  the  first  number.  What 
is  the  number? 

131.  A  farmer  bought  30  sheep  and  10  calves  for  the  same 
sum.  If  the  sheep  had  cost  25  per  cent  more  and  the  calves 
35  per  cent  less,  7  sheep  would  have  cost  $3  more  than  4 
calves.     What  did  each  sheep  cost? 

132.  Upon  withdrawing  from  the  business  A  takes  J  of 
the  capital  and  $100  more,  B  ^  of  the  new  remainder  and 
$100  more;  0  gets  $300.     What  was  the  capital? 

133.  What  number  is  that  which  gives  the  same  continued 
product  when  divided  into  3  equal  parts  as  when  divided  into 

4  equal  parts? 

134.  Find  a  number  of  two  digits,  the  first  of  which  is  4 
times  the  second,  and  the  number  is  2  less  than  3  times  the 
number  formed  by  inverting  the  digits. 

135.  In  going  from  one  town  to  another  a  traveller  found 
at  a  certain  place  that  the  distance  travelled  was  -|  the  whole 
distance,  and  when  he  had  gone  11  miles  further  he  had  f  of 
the  whole  distance  yet  to  go.     What  was  the  distance? 

136.  A  wine-merchant  has  wine  in  casks  of  two  sizes.  One 
containing  2|^  gallons   he  charges   $8.50  for;    the  other,  3^ 


PROBLEMS.  496 

gallons,  is  priced  at  $10.90.     Wliat  is  the  price  of  the  casks, 
supposing  them  to  cost  the  same? 

137.  A  miin's  income  was  $800  the  first  year,  and  increased 
$50  each  succeeding  year.  Afc  the  end  of  3  years  he  had 
L.ived  $15.75.      What  were  his  annual  expenses? 

138.  If  A  gives  B  $10  he  will  liave  twice  as  much  as  B; 
but  if  B  gives  A  $10  he  will  have  '^  as  much  as  A.  How  much 
had  each? 

(§  140.) 

PROBLEMS      INVOLVING      EQUATIONS      WITH       TWO      OR 
MORE     UNKNOWN     QUANTITIES. 

1.  It  is  found  that  when  a  ship  steams  12  knots  (sea-miles) 
an  hour  with  the  waves  she  pitches  1  in  15  seconds,  and 
steaming  at  the  same  speed  against  them  she  pitches  1  in  G 
seconds.  What  is  the  speed  of  the  waves,  and  how  many 
w Jives  are  there  in  a  sea-mile? 

2.  Two  men  start  at  the  same  time  to  miike  the  same 
journey.  The  first  goes  10  miles  the  first  day,  and  goes  a  cer- 
tain fixed  distance  more  every  following  day  than  he  did  the 
day  before.  He  overtakes  the  second  at  the  end  of  the  8th 
day,, and  finishes  his  journey  at  the  end  of  the  11th,  while  the 
second  finished  at  the  end  of  the  12th.  What  is  the  length 
of  the  journey,  and  how  far  did  the  second  go  each  day? 

3.  A  cannon  being  fired  while  a  heavy  wind  was  blowing, 
it  was  found  that  the  sound  required  4|-  sec(>nds  to  go  a  mile 
with  the  wind,  and  4|-  seconds  to  go  a  mile  against  the  wind. 
What  was  the  velocity  of  the  wind,  and  what  time  would  have 
been  required  for  the  sound  to  go  a  mile  in  still  air? 

4.  The  greatest  distance  between  Venus  and  the  earth  ij 
160  millions  of  miles;  the  least,  22  millions.  What  is  the  dis- 
tance of  each  from  the  sun,  supposing  that  each  moves  around 
the  sun  in  a  circular  orbit  having  the  sun  in  its  centre? 

5.  A  brother  and  sister  being  asked  how  large  the  family 
was,  the  brother  replied,  "1  have  as  many  brothers  as 
sisters."  The  sister  replied,  ^'  I  have  twice  as  many  brothers 
as  sisters."     How  many  boys  and  girls  were  in  the  family? 


496  SIMPLE  EQUATIONS. 

6.  Find  that  fraction  whose  value  becomes  ^  when  n  is  sub« 
tracted  from  each  of  its  terms,  and  ^  when  m  is  added  to  each 
of  its  terms. 

7.  Find  two  nunit)ers  such  that  their  difference  is  153, 
and  the  lesser  goes  into  the  greater  9  times  and  1  over. 

8.  One  number  divided  by  another  gives  the  quotient  4, 
with  3  as  a  remainder.  Increasing  divisor  and  dividend  by 
10,  the  quotient  is  2  and  the  remainder  23.   Find  the  numbers. 

9.  Find  two  quantities  such  that  half  their  sum  added  to 
half  their  difference  shall  make  a,  and  half  their  difference 
subtracted  from  half  their  sum  shall  leave  the  remainder  h, 

10.  Find  two  quantities  whose  sum  and  quotient  are  each 
equal  to  m. 

11.  Find  two  numbers  of  three  digits  of  which  one  is 
formed  by  simply  reversing  the  order  of  digits  in  the  other, 
and  which  fulfil  the  following  conditions:  (1)  the  sum  of  the 
digits  in  each  is  15;  (2)  the  sum  of  the  first  and  last  digits  is 
3  greater  than  the  second  one;  (3)  the  difference  of  the  num- 
bers is  99. 

12.  Each  of  two  vessels,  A  and  B,  was  partly  filled  with 
water.  A  man  poured  from  A  into  B  as  much  ms  was  already 
in  B,  then  from  B  into  A  as  much  as  was  left  in  A,  then  from 
A  into  B  as  much  as  was  left  in  B,  when  each  vessel  con- 
tained 8  quarts  of  water.    How  much  did  each  contain  at  first? 

13.  Find  two  quantities  the  sum  of  whose  reciprocals  is  5, 
and  ^  til  J  one  added  to  \  the  other  is  equal  to  twice  their 
product.* 

14.  For  $6.60  one  can  buy  either  20  pounds  of  coffee  and 
25  of  sugar  or  14  of  coffee  and  34  of  sugar.  What  is  the 
price  of  ejich  per  pound? 

15.  A  river  stermier  can  run  90  miles  down  stream  and 
back  again  in  15  hours;  but  if  she  runs  120  miles  down, 
she  can  only  get  back  70  miles  on  her  return  journey  at  the 
end  of  15  hours.  What  is  her  speed  and  the  flow  of  the 
river?  * 

16.  Cn  a  river  were  two  steamers,  the  speed  of  the  swift 

*  Compare  with  Exercises  11  to  20,  §  138. 


FEOBLBiMS.  497 

one  being  3  miles  an  hour  greater  than  that  of  tlie  slow  one. 
A  man  who  went  58  miles  down  the  river  on  the  slow  boat 
and  30  miles  back  on  the  swift  one  found  that  he  had  been 
9  hours  on  the  water.  But  when  lie  went  87  miles  down  on 
the  slow  boat  and  90  miles  back  on  the  swift  one,  he  found 
that  it  took  18  hours.  What  was  the  speed  of  each  boat  and 
the  flow  of  the  river? 

17.  A  quadrilateral  has  four  sides,  a,  I,  c  and  d.  If  \  of 
a  be  added  to  h,  then  \  of  the  extended  J)  be  added  to  c,  and 
then  ^  of  the  extended  c  to  d^  the  four  sides  will  each  be 
equal  to  m.     What  was  the  length  of  each  side  at  first? 

18.  Three  pedestrians  started  on  a  journey.  The  first 
performed  it  in  a  certain  time;  the  second,  going  1  mile  an 
hour  slower,  took  12  hours  longer;  the  third,  going  2  miles 
an  hour  slower  than  the  first,  took  33  hours  longer.  What 
was  the  distance,  and  the  speed  of  each? 

19.  The  perimeter  of  a  triangle  whose  sides  are  a,  h,  c,  is 
771  feet.  If  I"  the  side  a  be  added  to  b,  then  ^  of  the  prolonged 
h  be  added  to  c,  and  then  \  of  the  prolonged  c  be  added  to  a, 
the  sides  will  be  equal.     What  is  the  length  of  each  side? 

20.  Divide  232  into  three  parts.  A,  B  and  C,  such  that, 
whether  we  subtract  A  from  the  sum  of  B  and  C,  B  from  ^ 
the  sum  of  A  and  C,  or  C  from  J  the  sum  of  A  and  B,  the 
remainders  shall  all  be  equal. 

21.  Find  two  quantities  whose  difference  and  product  are 
each  equal  to  n, 

22.  The  quotient  of  two  numbers  is  2,  and  2  times  their 
sum  is  equal  to  6  times  their  difference.  What  are  the  num- 
bers? 

23.  A  man  has  a  saddle,  worth  $50,  and  two  horses.  If  the 
c:idule  be  put  on  horse  A,  he  will  equal  B  in  value;  but  if  put 
on  B,  his  value  will  be  double  that  of  A.  What  is  the  value 
of  each  horse? 

24.  What  number  of  two  digits  is  equal  to  4  times  their 
sum  and  12  times  their  difference? 

25.  What  number  of  two  digits  is  equal  to  4  times  their 
sum,  and  when  the  digits  are  reversed  equal  to  7  times  their 
Gum? 


498  SIMPLE  EQUATIONS. 

26.  Find  a  number  of  two  digits  that  is  equal  to  4  times 
the  Slim  of  its  digits  increased  by  3,  and  if  9  be  added  to  the 
number  the  digits  will  be  reversed. 

27.  Find  a  number  which  is  greater  by  2  than  6  times  the 
sum  of  its  digits,  and  if  9  be  subtracted  from  the  number  the 
digits  will  be  reversed. 

28.  What  number  is  that  which  is  4  times  the  sum  of  its 
digits,  and  is  3  greater  than  11  times  their  difference? 

29.  What  fraction  is  that  which  becomes  \  when  2  is  added 
to  the  denominator,  and  ^  if  5  be  subtracted  from  the  numer- 
ator? 

30.  Two  drovers  went  to  market  with  sheep.  A  sold  90 
and  then  had  left  ^  as  many  as  B.  Then  B  sold  72,  and  had 
f  as  many  as  A  remaining.     How  many  did  each  have? 

31.  A  woman  bought  60  apples  for  a  dollar,  giving  3  cents 
for  every  2  bad  ones  and  2  cents  each  for  the  good  ones.  How 
many  of  each  did  she  buy? 

32.  Find  a  fraction  that  becomes  ^  when  4  is  added  to  its 
denominator,  or  2  subtracted  from  its  numerator. 

T,T^,  A  marketman  had  4  more  ducks  than  chickens.  He 
sold  the  chickens  for  30  cents  apiece  and  the  ducks  for  40 
cents  apiece,  gaining  40  cents  more  than  if  the  prices  had 
been  reversed.     How  many  of  each  had  he? 

34.  A  boy  bouglit  a  number  of  apples  at  2  cents  each  and 
peaches  at  3  cents  each,  paying  $4.36  for  the  wliole;  12  of  the 
apples  were  bad  and  9  peaches  were  rotten.  He  sold  the  good 
apples  at  2  for  5  cents  and  the  peaches  3  for  10  cents,  receiv- 
ing $4.50  for  the  whole.     How  many  of  each  fruit  did  he  buy? 

35.  When  I  was  married  I  was  ^  older  than  my  wife;  10 
years  after  her  age  was  f  of  mine.  What  were  our  ages  when 
we  were  married? 

:^6.  A  and  B  can  do  a  piece  of  work  in  12  days;  but  if  A 
worked  twice  as  fast  they  could  do  it  in  8^-  days.  In  what 
time  could  eacli  of  thorn  do  it  singly? 

37.  B  and  C  can  do  a  piece  of  work  in  12  days;  with  the 
assistance  of  A  they  can  do  it  in  9  days.  In  what  time  can  A 
do  it  alone? 

38.  A  farmer  sold   60  fowls,  a  part  tujkcys  and  a  part 


PROBLEMS.  499 

chickens;  for  tke  turkeys  he  received  $1.10  apiece,  and  for 
the  chickens  50  cents  apiece,  receiving  for  the  whole  $51.60. 
How  many  were  there  of  each? 

39.  A  tank  has  4  pipes,  A,  B,  C  and  D.  A,  B  and  C  can 
fill  it  in  6  hours;  B,  C  and  D,  in  8  hours;  C,  D  and  A,  in  10 
hours;  D,  A  and  B,  in  12  hours.  How  long  will  it  take  each 
and  all  to  fill  it?    Explain  the  negative  result  for  D. 

40.  A  tank  has  two  pipes,  of  which  one  may  be  made  to 
run  either  in  or  out.  If  both  run  in  the  tank  is  filled  in  2 
hours;  if  one  in  and  the  other  out,  in  5  hours.  In  what 
times  would  the  separate  pipes  fill  it? 

41.  A  grocer  bought  50  pounds  of  sugar  and  100  pounds 
of  coffee  for  $26.  He  sold  the  sugar  at  an  advance  of  25  per 
cent  and  the  coffee  at  a  discount  of  10  per  cent,  receiving 
$25.50  for  the  whole.  What  was  the  buying  and  selling  price 
of  each? 

42.  Find  the  sum  of  two  numbers  the  difference  of  whose 
squares  is  equal  to  the  difference  of  the  numbers. 

43.  Divide  168  into  three  such  parts  that  the  second  divi- 
ded by  the  first  gives  5  as  a  quotient  and  10  for  a  remainder, 
and  the  difference  between  the  third  and  second  multiplied 
by  3  is  equal  to  4  times  the  first. 

44.  A  father  is  5  times  as  old  as  his  son.  Six  years  hence 
he  will  be  only  3  times  as  old.     What  are  their  present  ages? 

45.  The  sum  of  the  ages  of  two  persons  is  f  of  what  it 
will  be  12  years  hence.  The  difference  between  their  ages 
is  ^  of  w^hat  it  will  be  24  years  hence.  What  are  their 
ages? 

46.  A  farmer  sold  to  one  person  40  bushels  of  oats  and  30 
bushels  of  wheat  for  $44.50,  and  to  another  the  same  amount 
of  oats,  at  10  cents  a  bushel  more,  and  wheat,  at  5  cents  a 
bushel  less,  for  $57.     What  was  the  price  per  bushel  of  each? 

47.  There  is  a  number  of  3  digits  whose  sum  is  10.  The 
first  and  second  is  4  times  the  third,  and  if  29?  be  added  the 
digits  will  be  reversed.     What  is  the  number? 

48.  There  is  a  number  of  3  digits  whose  first  and  third 
digits  are  6  more  than  the  second.  Four  times  the  first  is 
14 more  than  the  difference  between  the  second  and  third;  and 


500  SIMPLE  EQUATIONS. 

if  97  be  added  to  the   number  the  digits  will  be  reversed. 
What  is  the  number? 

49.  A  certain  number  of  3  digits  is  34  times  the  sum  of 
its  digits,  and  also  102  times  the  difference  between  the  first 
and  second;  and  if  36  be  added  to  the  number  the  second  and 
third  will  exchange  places.     What  is  the  number? 

50.  An  oarsman  who  can  row  20  miles  and  back  in  7 
hours  finds  that  he  can  row  10  miles  with  the  current  in  the 
same  time  that  it  takes  him  to  go  4  miles  in  the  contrary 
direction.     Find  the  rate  of  the  current. 

51.  A  merchant  has  two  kinds  of  sugar;  one  cost  8  cents  a 
pound,  and  the  other  11  cents.  How  much  of  each  must  be 
taken  to  make  120  pounds  worth  9  cents  per  pound? 

52.  A  grocer  mixed  tea  that  cost  $1.10  a  pound  with  tea 
that  cost  95  cents  per  pound.  The  cost  of  the  mixture  is 
$101.  He  sells  it  at  $1  a  pound  and  gains  $2.  How  many 
pounds  of  each  did  he  use? 

53.  A,  B  and  C  can  earn  $25  in  5  days;  B  and  C,  $28  in  7 
days;  A  and  C,  $22  in  8  days.  What  does  each  man  earn  in  1 
day? 

^4.  A  and  B  can  do  a  piece  of  work  in  2  days;  A  and  C,  4 
times  as  much  in  9  days;  A,  B  and  C,  11  times  as  much  in  18 
days.     In  how  many  days  could  each  do  it  alone? 

55.  A  sum  of  money  at  simple  interest  amounted  in  5  yeais 
to  $1500,  and  in  8  years  to  $1680.  What  was  the  principal 
and  rate? 

56.  A  ]icrson  has  $1200  invested  at  a  certain  rate  and  for 
a  certain  time;  had  the  rate  been  1  per  cent  less  and  the  time 
2  years  more,  he  would  have  had  $24  more  interest;  while 
with  a  rate  2  per  cent  less  and  a  time  1  year  more  he  would 
have  had  $144  less  interest.     Eind  the  rate  and  time. 

57.  A  sum  of  money  at  simple  interest  for  c  years 
amounted  to  t  dollars,  and  the  same  for  h  years  amounted  to 
a  dollars.     What  was  the  principal  and  rate? 

58.  In  a  race  over  a  course  4000  feet  long  A  gives  B  300 
feet  start,  and  wins  by  1  minute  and  20  seconds.  In  a  second 
trial  A  gives  him  40  seconds  start,  and  wins  by  900  feet. 
What  was  the  rate  of  each? 


FllOBLEMS.  501 

59.  A,  B  and  C  promised  to  give  $1000  to  a  church.  A  gave 
one  third  less  than  he  agreed  to,  so  B  increased  his  by  one 
fourth,  which  left  $55  more  for  C.  Now  if  B  had  given  one 
fifth  less  than  promised,  and  C  $70  more,  A's  share  would 
have  been  his  original  subscription.  What  was  the  amount 
of  the  first  pledge? 

60.  The  fore  wheels  of  a  carriage  are  10^  feet  in  circum- 
tVreiice,  and  the  hind  wheels  13.  in  going  a  journey  tlie 
fore  wheels  make  2500  more  revolutions  than  the  hind  wheels. 
What  was  the  distance? 

61.  A  coach  has  2  more  outside  passengers  than  inside. 
Six  outsiders  could  travel  at  an  expense  of  $1  more  than  4  in- 
siders. The  fare  of  all  amounted  to  $20.50.  At  the  end  of 
half  the  journey  2  were  added  to  the  outside  and  1  inside, 
Avhich  increased  the  total  fare  by  $2.50.  What  was  the  num- 
ber and  fare  of  each  class? 

62.  A  person  has  two  creditors;  at  one  time  he  ])ays  them 
$680,  giving  to  one  f  of  the  sum  due  him,  and  to  the  other 
$40  more  than  ^  of  his  debt;  at  another  time  he  pays  them 
$580,  giving  to  the  first  f  of  what  remains  due  to  him,  and  to 
the  other  -f-  of  what  remains  due  to  him.  Wiiat  was  the 
amount  of  each  debt?- 

63.  If  a  certain  croquet-ground  were  5  feet  longer  and  3 
feet  broader  it  would  contain  320  more  feet;  but  if  it  were  3 
feet  longer  and  5  feet  broader  it  would  contain  310  more  feet. 
What  is  its  present  area? 

64.  The  sum  of  two  numbers  is  12,  and  the  difference  of 
their  squares  is  24.     What  are  the  numbers? 

65.  Two  boats,  320  and  360  feet  long  respectively,  are 
moving  with  uniform  speed.  If  they  go  in  opposite  directions 
it  requires  10  seconds  to  pass  each  other;  but  if  they  go  in  the 
same  direction  it  takes  90  seconds  for  them  to  pass.  What  is 
the  speed  of  each  boat? 

66.  A  train  runs  a  certain  distance  at  a  uniform  rate.  If  the 
rate  be  increased  by  5  miles  an  hour  the  distance  would  be 
travelled  in  f  of  the  time;  but  if  the  rate  be  diminished  by  5 
miles  an  hour  the  time  Avould  be  increased  by  3  hours.  Wliat 
is  the  rate  and  distance? 


502  SIMPLE  EQUATIONS. 

67.  What  number  of  3  digits  is  greater  by  99  when  its 
digits  are  reversed;  greater  by  270  than  the  sum  of  its  digits; 
and  greater  by  45  than  when  the  second  and  third  are  trans- 
posed ? 

6^,  A  and  B  could  have  completed  a  certain  piece  of  work 
in  12  days;  but  after  both  had  worked  4  days  B  was  left  to 
finish  it  alone,  which  he  did  in  24  days  more.  How  long 
would  it  have  taken  each  to  do  it  alone? 

69.  A  number  consists  of  2  digits  whose  sum  is  12,  and  if  15 
be  subtracted  from  the  number,  and  the  remainder  be  divided 
by  2,  the  digits  will  be  inverted.     What  is  the  number? 

70.  A  boy  spent  his  money  in  oranges.  If  he  had  bouglit 
5  more,  each  orange  would  have  cost  a  half-cent  less;  if  3  less, 
a  half-cent  more.  How  much  did  he  spend,  and  how  many 
did  he  buy? 

71.  A  person  bought  apples  at  4  cents  a  dozen,  and  1| 
times  as  many  peaches  at  12  cents  a  dozen;  after  mixing  them 
he  sold  them  at  8  cents  a  dozen,  losing  4  cents  on  the  whole. 
How  many  dozen  of  each  did  he  buy? 

72.  Fi»nd  a  fraction  that  becomes  f  when  2  is  added  to 
its  numerator,  and  ^  when  4  is  added  to  the  denominator. 

73.  Five  pounds  of  tea  and  12  pounds  of  sugar  cost  $7.44. 
If  tea  were  to  rise  10  per  cent  and  sugar  fall  25  pe:  cent,  8 
pounds  of  tea  and  6  pounds  of  sugar  would  cost  $11.10.  What 
is  the  price  per  pound  of  each? 

74.  A^s  income  is  half  as  much  again  as  B%  while  his  ex- 
penses are  twice  as  great  as  B's.  A  spends  $60  more  than  his 
income,  and  B  $60  less  than  his.   What  is  the  income  of  each? 

75.  A  invested  some  money  at  5  per  cent,  and  B  at  6  per 
cent,  both  receiving  the  same  amount  of  income.  If  A  had 
invested  $1000  more  than  he  did,  his  income  would  have  been 
11  per  cent  on  B's  investment.     What  did  each  invest? 

76.  An  oarsman  can  row  9  miles  up  stream  and  13  miles 
down  in  4  hours,  or  13  miles  up  and  "9  miles  down  in  5  hours. 
What  is  the  rate  of  the  stream  and  of  the  rowing? 

77.  Six  years  hence  the  product  of  two  people's  ages  will 
be  greater  by  348  than  it  is  now.  What  will  then  be  the  sum 
of  their  ages? 


PROBLEMS,  503 

78.  A  invests  money  at  4  per  cent,  B  at  5  per  cent,  and  C 
at  6  per  cent.  A  and  B  together  receive  ^560,  B  and  C  $520, 
and  A  and  C  $360.     How  much  does  eacli  invest? 

79.  Find  the  quotient  of  two  numbers  whose  sum  is  n 
times  their  difference. 

80.  A  and  B  can  finish  a  job  in  12  days.  A  worked  2  days, 
and  B  3.  How  long  will  it  take  C  to  finish  it  if  he  could  have 
done  the  whole  in  15  days  with  B^s  assistance,  and  in  10  days 
with  A's? 

81.  A  cai'penter  and  apprentice  received  $16.80  for  7  days' 
wages,  the  carpenter  getting  20  cents  more  for  2  days'  work 
than  the  boy  for  3  days'.     What  was  the  daily  wages  of  each? 

82.  A  man  paid  $50  for  7  photographs  and  12  prints;  if 
he  had  paid  $1  more  he  could  have  had  7  prints  and  15 
photographs.     What  was  the  price  of  each? 

Ratio  and  Proportion. 
§164. 

1.  Divide  126  into  three  parts  that  shall  be  proportional 
to  the  numbers  3,  4,  7. 

2.  Find  two  fractions  that  shall  be  to  each  other  as  3  :  4, 
and  whose  sum  shall  be  f.  • 

3.  Divide  .0444  into  three  parts  that  shall  be  to  each  other 
as  i  :  i  :  |. 

4.  Find  two  numbers  which  are  to  each  other  as  4  :  3,  and 
whose  difference  is  ^  of  the  less  ? 

5  If  0-  :  ^  ::  6  :  8  and  4a;  —  3^  =  7,  what  is  the  value  of 
X  and  y'^ 

6.  A  year's  profits  were  divided  among  two  partners  in  the 
proportion  of  3  :  4.  If  the  second  should  give  $425  to  the 
first,  their  shares  would  be  equal.  What  was  the  amount 
divided? 

7.  In  a  first  yearns  partnership  A  had  3  shares,  and  B  4. 
In  the  second,  A  had  1,  and  B  2.  In  the  first  year  A  gained 
$300  more  than  he  did  the  second,  and  B  gained  $200  less 
than  he  did  the  second.     What  were  the  profits  each  year  ? 


504  RATIO  AND  PROPORTION. 

8.  In  a  farm-yard  there  are  4  sheep  to  every  3  cattle,  and 
5  cattle  to  6  hogs.  How  many  hogs  are  there  to  every  20 
sheep? 

9.  A  drover  started  to  market  with  a  herd  of  7  horses  to 
every  5  mules.  He  sold  27  horses  and  bought  3  mules,  and 
then  had  3  horses  to  every  4  mules.  How  many  of  each  had 
he  at  first? 

10.  Find  two  quantities  whose  sum,  difference  and  product 
are  proportional  to  5,  1  and  12. 

11.  What  number  is  that  to  which  if  2,  6  and  12  be  sever- 
ally added,  the  first  sum  shall  be  to  the  second  as  the  second 
is  to  the  third? 

12.  What  two  numbers  are  to  each  other  as  3  to  4,  and  if 
4  be  added  to  each  the  sums  will  be  as  4  to  5? 

13.  What  quantity  must  be  taken  from  each  term  of  the 
ratio  m  :  n  that  it  may  equal  the  ratio  c  \  d? 

14.  If  a  :  Z>  be  the  square  of  the  ratio  of  a  -\-  c  :  b  -\-  c, 
show  that  c  is  a  mean  proportional  between  a  and  b, 

15.  If  a  :  Z>  ==  Z>. :  c,  show  that  a:a-{-b=^a  —  b\a  —  c, 

16.  And  under  the  same  conditions  show  that 

{a'  +  ¥)  {V  +  c')  =  {ab  +  bc)\ 

17.  li  a  :  b  —  c  :  d,  show  that 

a  {a  -{-  b  -{-  c  -{-  d)  =  {a'\-b)  {a  -\-  c). 

18.  In  a  milk-cUn,  the  quantity  of  milk  is  to  the  entire 
contents  (milk  and  water)  as  5  :  6.  Five  gallons  are  sold,  and 
1  gallon  of  water  is  added;  then  the  ratio  of  the  milk  to  the 
whole  is  4  :  5.    How  many  gallons  of  each  were  there  at  first? 

19.  In  a  two-mile  race  between  a  bicycle  and  a  horse,  their 
rates  were  as  5  to  6.  The  bicycle  had  1  minute  start,  but 
was  beaten  by  312  yards.     What  was  the  rate  of  each? 

20.  A  line  is  divided  by  one  point  into  two  parts  in  the 
ratio  of  3  :  5,  and  by  another  point  into  two  parts  in  the  ratio 
of  1  :  3.  The  distance  between  the  points  of  division  is  1 
inch.     What  is  the  length  of  the  line? 

21.  The  sum  of  the  two  digits  of  a  number  is  6,  and  the 
number  is  to  the  number  expressed  by  the  same  digits  reversed 
as  4  :  7.     What  is  the  number? 

22.  One  ingot  contains  two  parts  of  gold  and  one  of  silver, 


PROBLEMS.  505 

and  anotlier  two  parts  of  gold  and  three  of  silver.  If  equal 
parts  are  taken  from  each  ingot,  what  will  be  the  proportion 
of  the  gold  to  the  silver  in  the  alloy? 

23.  If  two  ounces  be  taken  from  the  first  and  three  from 
the  second,  what  will  be  the  ratio  of  the  gold  to  the  silver? 

24.  A  cask  contains  4  gallons  of  water  and  18  gallons  of 
alcohol.  How  many  gallons  of  a  mixture  containing  2  parts 
water  and  5  parts  alcohol  must  be  put  in  the  cask  so  that  there 
may  be  2  parts  of  water  to  7  of  alcohol? 

25.  Which  is  the  greater  I'atio,  1  -\-  a  \  1  —  a  or  1 -\-  a^  : 
1  —  a^,  a  being  positive  and  less  ihan  1? 

26.  AVhicli  is  the  greater  ratio,  a^—  ah  -{-If  :  a^  -\-  ab  -\-  F 
or  a*  —  a^'b''  +  ^*  :  a*  +  a^b^  +  ^\  ^  '^^^  ^  having  like  signs? 

27.  What  number  must  be  taken  from  the  second  term  of 
the  ratio  2  :  34  and  added  to  the  first  that  it  may  equal  5:6? 

28.  What  number  must  be  taken  from  each  term  of  the 
ratio  19  :  30  that  it  may  equal  the  ratio  1:2? 

29.  It  a  :b  =  c  :  d,  show  that  a^  :  b""  =  a"  -\-  c"  :  F  +  d\ 

30.  A  bankrupt  owed  two  creditors  $1800.  The  sum  of 
their  credits  is  to  the  less  as  3  :  1.     What  did  he  owe  each? 

31.  Discuss  the  general  problem:  To  divide  a  given  quantity 
iVinto  parts  proportional  to  the  given  numbers  m,  n,  p,  etc. 

32.  Divide  the  number  iV^into  three  parts,  x,  y  and  z,  such 
that  X  shall  be  to  t/  as  2  :  3,  and  z  to  the  difference  between  x 
and  y  as  3  :  2. 

2,2,'  The  speed  of  the  steamship  Servia  is  to  that  of  the  Both- 
nia as  13  to  10,  and  the  first  steams  5  miles  farther  in  8  hours 
than  the  second  does  in  10  hours.     What  is  the  speed  of  each? 

34.  The  speed  of  two  pedestrians  was  as  4:3,  and  the 
slower  was  5  hours  longer  in  going  36  miles  than  the  faster 
was  in  going  24.     What  was  the  rate  of  each? 

35.  A  chemist  had  two  vessels,  A,  containing  acid,  and  B,  an 
equal  quantity  of  water.  He  poured  one  third  the  acid  into 
tlie  water,. and  then  poured  one  third  of  this  mixture  back  into 
the  acid.     What  was  then  the  ratio  of  acid  to  water  in  A? 

2)6.  If  24  grains  of  gold  and  400  grains  of  silver  are  each 
worth  one  dollar,  what  will  be  the  weight  of  a  coin  containing 
equal  parts  of  gold  and  silver  and  Avorth  a  dollar? 


506  llATIO  AND  PROPORTION 

37.  AVluit  common  quantity  must  be  subtracted  from  the 
four  quantities  m,  n,  x  and  y  that  the  remainders  may  form 
a  proportion? 

2,^.  A  cliemist  has  two  mixtures  of  alcohol  and  water,  the 
one  containing  90  per  cent,  of  alcohol,  the  other  50  per  cent. 
How  much  of  the  first  must  he  add  to  1  litre  of  the  second  to 
make  a  mixture  containing  80  per  cent,  of  alcohol? 

39.  It  is  a  law  of  mechanics  that  the  distances  through 
which  heavy  bodies  will  fall  in  a  vacuum  in  different  times  are 
23roportional  to  the  squares  of  the  times.  If  a  body  fall  48 
feet  fai'ther  in  2  seconds  than  in  1  second,  how  far  will  it  fall 
in  1  second?     How  far  in  t  seconds? 

40.  Find  an  expression  snch  that  if  yon  subtract  m  +  n 
and  711  —  71,  the  ratio  of  the  remainders  shall  be  n  :  m. 

41.  On  a  line  are  two  points  whose  distance  is  a.  The 
first  point  divides  the  line  into  parts  whose  ratio  is  2  :  3;  the 
second  into  parts  whose  ratio  is  5  :  7.  What  is  the  length  of 
the  line? 

42.  If  a  line  is  divided  into  two  ])arts  whose  ratio  is  m  :  n, 
what  is  the  ratio  of  the  length  of  the  whole  line  to  the  distance 
of  the  point  of  division  from  the  middle  point? 

43.  A  line  is  divided  into  three  segments  propoitional  to 
the  numbers  m,  p  and  q.  What  is  the  ratio  of  the  parts  into 
which  the  middle  point  of  the  line  divides  the  middle  segment? 

44.  Divide  $285  among  tlii'ee  persons,  A,  B  and  C,  so  that 
the  share  of  A  shall  be  to  that  of  B  as  6  .  11,  and  that  of  C 
shall  be  $30  more  than  those  of  A  and  B  together. 

45.  A  sailing-ship  leaves  port,  and  12  hours  later  is  fol- 
lowed by  a  steamship.  If  the  ratio  of  the  speeds  is  3  :  8 
how  long  will  it  take  the  steamer  to  overtake  the  ship? 

46.  A  courier  started  from  his  post,  going  7  miles  in  3 
hours.  Two  hours  later  another  follow'ed,  going  7  miles  in  2 
hours.     How  long  will  the  second  be  overtaking  the  first? 

47.  The  areas  of  the  openings  of  two  water-faucets  are  in 
the  ratio  3  :  5;  the  speeds,  of  flow  of  the  water  through  the 
openings  are  in  the  ratio  3:4.  At  the  end  of  an  hour  1221 
gallons  more  have  flowed  through  the  second  than  through  the 
first.     What  was  the  flow  from  each? 


PROBLEMS,  507 

48.  Tlie  flows  from  two  faucets  into  two  equal  vessels  is  in 
the  ratio  4  :  7,  and  both  vessels  were  placed  under  them  at 
the  same  moment.  When  the  vessel  under  the  larger  faucet 
was  full,  it  was  removed  and  the  other  put  into  its  place.  In 
80  seconds  from  the  time  of  beginning  both  vessels  were  filled. 
How  long  would  it  take  each  f;iucet  to  fill  one  of  the  vessels? 

49.  Three  numbers,  a,  b  and  c,  are  so  related  that 

a  :  h  -\-  c  —  m  :  n, 
b  :  c  -\-  a  =  p   :  q. 

Find  the  ratio  c  :  a  -{-  b.     Find  a,  h,  and  then  a  -\-  h,  in  terms 
of  c, 

50.  If,  in  the  preceding  problem,  the  &\\m  a -{- h  -\-  c  =^  X, 
express  each  of  the  numbers  a,  h  and  c  in  terms  of  N. 

51.  The  speeds  of  two  trains,  A  and  B,  are  as  m  :  ti,  and 
the  journeys  they  have  to  make  as  ^  :  5'.  It  took  train  B  t 
hours  longer  to  make  its  journey  than  it  did  train  A.  What 
was  the  time  required  by  each  train  for  each  journey? 

52.  A  street-railway  runs  along  a  regular  incline,  in  conse- 
quence of  which  the  s})eeds  of  the  cars  going  in  the  two  direc- 
tions are  as  2:3.  The  cars  leave  each  terminus  at  regular 
intervals  of  5  minutes.  At  what  intervals  of  time  will  a  car 
going  up  hill  meet  the  successive  cars  coming  down,  awdivicG 
versa'^ 

53.  The  same  thing  being  supposed,  two  cars  starting  out 
simultaneously  from  the  termini  meet  at  the  end  of  30  minutes. 
How  long  in  time  is  the  journey  for  each  car? 

54.  The  same  thing  being  again  supposed,  a  rider  gallops 
up  hill  at  such  a  rate  that  he  passes  the  successive  cars  going 
up  hill  at  the  same  time  that  they  meet  the  successive  cars 
coming  down,  so  that  every  time  he  passes  a  car  going  up  he 
meets  one  coming  down.  What  is  the  ratio  of  his  speed  to 
that  of  each  of  the  cars? 

55.  Give  the  algebraic  answers  to  the  three  preceding 
questions  when  the  ratio  of  the  speeds  is  m  :  w. 

56.  Three  given  points.  A,  B  and  X,  lie  in  a  straight  line. 
A  and  B  are  taken  as  base-  ^  y  b  x 
points  from  which  distances     '                           ill 


508  RATIO  AND  PUOPOIITION. 

are  measured.  Having  given 

Distance  AB  =  b, 
Distance  AX  =  x, 

it  is  required  to  find  the  jDositioii  of  a  fourth  point,  Y,  between 
A  and  B,  such  that  we  shall  have 

AX  :YB  =  AX  :BX  =  X  :  X  -  h. 

Do  this  by  finding  the  distance  of  Y  from  A  in  terms  of  h 
and  X. 

57.   Show  that  in  the  preceding  construction  we  have 


AY   '    AX        AB* 

58.  Show  that,  in  the  preceding  problem,  the  product  of 
the  distances  of  X  and  Y  from  the  middle  point  of  the  line 
AB  is  \h\ 

59.  If,  instead  of  the  point  X,  the  point  Y  is  given,  find 
the  distances  AX  corresponding  to  the  following  values  of 
AY,  in  order  that  the  same  proportion  may  hold  true,  and 
explain  the  results  when  negative: 


(«)  AY  =  i  J. 

Ans.  X 

=  !»• 

{e)  KY  =  \  A. 

(^)AY  =  |s. 

(,?)  AY  =  ^  A. 

(;/)  AY  =  i  6. 

(V)  AY  =.  1  A. 

{d)  AY  =  (i  + 

«)^>. 

(e)AY  =  ^A. 

I1p:mark.  When  four  points  on  a  strnight  line  fulfil  the  preceding 
proportion,  they  arc  called  four  harmonic  points,  and  the  line  AB  is 
said  to  be  divided  harmonically. 

60.  It  is  a  theorem  of  mechanics  that,  in  order  that  two 
masses,  V  and  AV,  at  the  ends  of  a  lever,  AB,  may  be  In  equi- 
librium, the  distances  of  their  points  of  suspension,  A  and  B, 
from  the  fulcrum,  F,  must  be  inversely  j^voportional  to  their 
weights;  that  is,  we  must  have 

Weight  V  :  weight  W  =  FB  :  FA. 


PROBLEMS.  C09 

iNow,  if  the  leiigtli  AB  of  the  lever  is  I,  and  the  weights  of 
A F  B 


V  and  W  are  respectively  m  and  n,  express  the  lengths  AF 
and  FB  of  the  arms  of  the  lever. 

6 1.  The  weights  at  the  end«  of  a  lever  are  8  and  13  kilo- 
grammes, and  the  fulcrum  is  3  inches  from  the  middle  of  the 
lever.     What  is  the  length  of  the  lever? 

62.  The  sum  of  the  two  weights  is  25  pounds,  and  the 
ratio  of  the  distance  of  the  fulcrum  from  the  middle  point  to 
the  length  of  the  lever  is  2  :  9.     What  are  the  weights? 

6:^,  The  weights  are  m  and  n  {in  >  71),  and  one  arm  of  the 
lever  is  h  longer  than  the  other.  Express  the  length  of  the 
lever. 

64.  A  lever  was  balanced  with  weights  of  7  and  9  kilo- 
grammes at  its  ends.  One  kilogramme  being  taken  from  the 
lesser  and  added  to  the  greater  (making  the  weights  6  and  10 
kilogrammes),  the  fulcrum  had  to  be  moved  2  inches.  What 
was  the  length  of  the  lever? 

65.  A  line  is  divided  into  three  parts  proportional  to  the 
numbers  3,  4  and  5.  What  is  the  ratio  of  the  parts  in  which 
the  middle  point  of  the  line  divides  the  middle  segment? 

66.  To  300  pounds  of  a  mixture  containing  2  parts  of  zinc, 
3  of  copper  and  4  of  tin  was  added  200  pounds  of  another 
mixture  of  the  same  metals,  when  it  was  found  that  the  pro- 
portions were  now  as  3,  4  and  5.  What  Avere  the  proportions 
in  the  mixture  added? 

67.  Find  two  numbers  whose  sum,  difference  and  product 
are  to  each  other  as  the  numbers  5  :  1  :  18. 

6S,  Find  two  numbers  in  the  ratio  7  :  3,  the  ratio  of 
whoso  difference  to  their  product  is  1  :  21. 


510  •  RATIO  AND  PROPORTION, 

69.  Find  fcAvo  numbers  such  that  the  first  sliall  be  to  the 
second  as  their  sum  is  to  3^,  and  as  their  difference  is  to  2f . 

70.  Find  three  numbers  whose  sum  is  73,  and  such  tliat  if 
2  be  subtracted  from  the  first  and  second  their  differences  will 
be  to  each  other  as  1  :  2,  and  if  2  be  added  to  the  second  and 
third  their  sums  Avill  be  to  each  otlier  as  4  :  5. 

71.  Two  boats  start  in  a  race.  The  second  boat  rows  25 
strokes  to  the  first's  28,  but  10  strokes  of  the  second  are  equal 
to  12  of  the  first.  If  the  distance  between  the  boats  at  starting 
is  30  strokes  of  the  second  boat,  how  many  strokes  will  it: 
make  before  reaching  the  first? 

72.  One  cask  contains  18  gallons  of  wine  and  6  gallons  of 
water;  another  contains  12  gallons  of  wine  and  18  gallons  of 
water.  How  much  must  be  taken  from  each  to  form  a  mix- 
ture containing  8  gallons  of  wine  and  8  gallons  of  water? 

73.  Two  mixtures  of  wine  and  water  contain  respectively 
\  and  I  wine.  How  much  of  each  must  be  taken  to  form  44 
gallons  of  a  mixture  of  which  the  wine  is  to  the  water  as  5  :  6? 

74.  A  and  B  ran  a  race  in  6  minutes.  B  had  a  start  of 
20  yards;  but  A  ran  5  yards  while  B  ran  4,  and  Avon  by 
10  yards.  What  was  the  length  of  the  race,  and  the  rate  o'f 
running? 

75 .  A  jeweller  has  three  ingots  of  metal.  A  pound  of  the 
first  contains  7  ounces  of  gold,  3  ounces  of  silver  and  6 
ounces  of  copper;  a  pound  of  the  second  contains  12  ounces 
of  gold,  3  ounces  of  silver  and  1  ounce  of  copper;  a  i^ound 
of  the  third  contains  4  ounces  of  gold,  7  ounces  of  silver  and 
5  ounces  of  copper.  He  wishes  to  form  an  alloy  weighing  1 
pound,  which  shall  have  8  ounces  of  gold,  3f  ounces  of  silver 
and  ^\  ounces  of  copper.  How  much  must  be  taken  from 
each  ingot? 

76.  The  king  of  Syracuse  gave  a  goldsmiih  10  pounds  of 
gold  with  which  to  make  a  crown.  When  it  was  finished  the 
king  gave  (lie  crown  to  Archimedes  to  ascertain  if  it  was  pure 
gold.  The  philosopher  knew  that  gold  weighs  .948  as  much 
in  water  as  in  air,  and  silver  .901.  When  the  crown  was 
weighed  in  water  he  found  it  lost  10  ounces.  What  was  the 
quaniity  of  gold  and  silver  in  the  crown? 


IRRATIONAL  EXPRESSIONS.  gH 

Irrational  ExpressionSc 
§179, 

Execute  tlie  following  divisions: 

1 .  x^^  -f-  or^. 

2.  )llx^h^^ci  ~  9a^Wc. 

3.  12^Wi  -^4aW. 

4.  x^^^-yz^  -V-  x'^y^z~l. 

5.  a-'b-'-^a-'h-K 

6.  x-\i~^z~'^  -, x^yz^. 

7.  a-^"^-^c-^^-«w. 

9.  24:cy^  +  15:c-y;s'  -  '^xy-'  H-  B.^-y. 

10.  ^hx'yz-'  -  ^Wy-^z'  +  ly-'z'  ~  -  IxY^-'. 

1 1 .  20a; V -  'z'  -  4.y'z'  -12x-'y-'-^4:X-'y- 'z\ 

12.  28%^-'  +  l%xy-''z^  —  12x-^y-'z-'  -^  4x-hj-'z-\ 

13.  ai  —a^U  -^  «i 

14.  X^  —  xW  -\-X^  -^  Xh. 

15.  12at  -36at  -^  l-^ftl 

m 

16.  2a;«  -  60;-^-^  2a;'. 


17.  80;^  —4:xi  -^  2.T~i 


183. 


Express  the  following  ])rodacts  of  irrational  qnantities 
with  a  single  fractional  exponent  by  reducing  the  fractional 
exponents  to  a  common  denominator,  and  then  reversing  the 
process  of  §182: 

X  y  z  1 

I.   Prove  the  equation"  «w^«c«  =  {d^b^c^Y, 
2/  aWck     Ans.  a^bk'^  =  {a^b'c)i. 


3- 

m^n^pK 

4. 

2m.     All 

IS.  12i 

5. 

2i3i 

6. 

am. 

7. 

MrisK 

8. 

1    1 

x^y^^. 

9- 

d2iam. 

10. 

vrvvvi 

II. 

ami. 

12. 

cm. 

13. 

7^ .  5i 

14. 

h-m 

15- 

2-^3^. 

16. 

8^12- i 

512  IRRATIONAL  EXPRESSIONS. 

§183. 

Keduce  the  following  expressions  to  monomials: 

1.  Vbi+  V2i+  '/K       Alls.    ¥6{Vd+ V4t-]-i)=  oVif, 

2.  VT2  +  V27  +  i^i^.  3.    V6-  V80  +  VlTo. 
4.    V2-  Vis  +  V32,  5.    V76  +  4/48  -  VS, 

6.  1^2  +  2  ^27"+  3  fT5  -  9  1^48. 

7.  V4^  -  Vda  +  V2da.  8.    V'^-  +  Vb^  -  V7^. 
Eeduce  the  following  to  their  simplest  form  and  factor: 

9.  Vl^^F  +  VbUa^\  10.   {4:a'by  -  {a'dy. 

11.  {2''a''b'c)^  -  {4..b'a'b'c')i  +  {4..6'ah'c)l 

12.  {64.a'^  +  'F)h  -  {Ua'^-Wy^  +  {2a*"*  +  ^)i  +  {2a'^cy. 

13.  Kc^  +  ^W.  14.    [{ci  +  ^y{x  +  y)]K 


,5.  iL+i  l/^^ZJ.  x6. 


7)1  —  11  f  mp 


-  - 


q      2^ -{-  q  '   7ri-\-  h\ni'  —  2m)i-{-  if  J  ' 

§184o 
Multiply: 

I.   {c  +  b  V7)  {c  -  a  V7).       2.   (m  +  Vji)  {vi  -  VJi). 

3.  {am  -\-  n  Va  —  z)  {n  —  m  Va  —  z). 

4.  (4  _[_  3  |/2)  (4-3  V2),      5.   (5  -  Qn  V2)  (5  +  ijn  ^2). 

6.  Mi-(^  -  \m^  +  2(i^iw. 

7.  ( i^i? + ^ + '^/^  -  ^)  ( ^/^ + ^  -  ^^^  -  ^)- 

8.   {a  +  x^  +  yi){a~'0^-\-y^.^ 

I  Vvi  Vn  UVm    _    Vn\ 

^'   ^Vn  VmJ^Vn  VnJ' 

{x+a)^  \   \{x-aY 

{x  -  ay  \\{x+  ay  ^ 

Aggregate  the  following  fractional  expressions  and  sim- 
plify when  possible: 

ri    .    a^  {c-\-xy       (c-x)^ 

a        r  {c  —  xp       [c  -\-  xy 

ml     ,   m^    ,   m  ,    .     ^'  +  ^ 

'  ^3*   —T  -^ r  -^ — •  14.   7-  —  t  -\ ; . 


QUADRATIG  EQUATIONS.  513 


Equations  of  the  Second  Degree. 

§§  195-302. 

I.   ^x'  +  nx  =  1 1.0.  2.   2.r  +  8a;  =  64. 

3.  a;^  -  llx  +  6  ==  57.  4.  ^'  -  37a;  =  -  320. 

5.  a;'  +  6.T  ==:  7.  6.  a;'  -  %x  =  -  12. 

7.  x  —  mx  —  —  n,  Z.  X  ^  —  —  --. 

2         2^/' 

15       72  -  6a;       ^  ,  ,  ^ 

Q. --^ —  =  2.  10.  a;  —  ax  —  bx  :=  —  ao. 

X  2x 

II.    V'V  —  1  =  a;  —  1.  12.  a;^  —  of  a;  =  18. 

13.  3a;^  +  a;  =  7.  14.  4.i' ^^  ==  46. 

40      ,    27       ,,,  ^48  165 

19.  (a;  —  fl^)  (x  —  h)  =  0, 

20.  (a;  +  4)  (a;  +  1)  =  6  (x'  +  1)  -  8a;^ 

21.  3  (a;'  -  1)  -  24  =  4  (a:  +  5)  {x  -  3). 

22.  {x  -  2)  (3a;  +  1)  =  10  -  (2a;  +  1)  (^  _  3). 

X  a;  —  3  a;  —  8         x  —  1 


2  (a;  -  3)       x  -  I'  ^'  a;  +  2       2a;  +  10* 

5 29  _  _  3 

2  —  X       4  —  5a;       2a;* 

a;  +  2       _  3  16a; 

32-  (2a:  -  1)  ~"  ^  "^  4a;'  -  1* 

X  —  1       2a;  —  3       a;  —  8     ^        a  x  +  b 

2*7,  = ,  28. = ' — . 

'x  a;  —  1        a;  —  9'      '  x  —  b       %x  —  a 

3  2  1 


^5- 
26. 


29. 


?>a  —  a;       2«  —  a;       a  —  3a;* 


30.   6  |/.i;  +  -^  ==  37.  31.    |/3i^;  -  5  +  |/a;  +  6  3^  9. 

32.    \^x  —  2  +  1/4  —  a;  =   4/0  —  a;. 


614  >l'^2'^^   OA'E   UNKNOWN  QVANTJTT. 

a-x       9 --3.7;    ,    , 
33.   10  -^-  =  ----  +  00;. 


X 


+  3    ,16  ---  -^x  __  26 


35.   14  +  3.T  -  ^;^^----  =  yx  H ^ — . 

X  H-  19  _  4  _  9.7;  -  8 
^  *  '     3  a;  ~       ;<i      ' 

^        ,    ^T  +  1        13 


iz;  +  1    ^      a;  6  *  ^  ■.-?;  +  6 

^9-   --3 9"~  -  V^Z  ~  ^• 

,;  4.  11  9  +  4:^ 

4c. ==7 i — . 

x  X 

41.   (3.T  +  1 )  (4:t'  -  2)  ^  (13.T  +  7)  (5^-  -  3). 

;^  +  2a;^  +  8 
x'  4-  ^'  —  G 


42.    -X— -.^  ^x^^x-^^. 


43- 


(2:  -  ])  (:r  -  2)  +  (^  -  2)  (.T  -  4)  =  12^;  -  30. 
10 H)_    _       3  82:      _  ^    ,    20 

8  —  2:        2^—  ll_:r  —  2 
4  •    — ^  a;  -  3     ~       6     * 

3.T  -  3        ,      ,    3a;  -  6 

47.   ^---^— 3-  =  4^  +  -T— 
2:r  +  3    ,    J_  _  1    ,    -,  5a:  +  3        2.t  -  3  _ 
^°'   2a;  +  1  +  2a;^  ~  a;  +    •    ^'^     2;  -  1    +  2ar^-2  "  ^• 

52.  |/4"+V+ |/:^  r=  3. 

53.  2a:  -  a:'  +  1^0^^  —  12a:  +  7  =  0. 

54.  f'a:'!^  +  \^x  +  8  =  5  l^i. 

55.  4/2a^+  1  +  \/'lx-'ri  r=  |/3.i;  4-  4; 


QUADRATIC  EQUATIONS.  515 

3a;  —  5       5a;  —  3 
2^  4  6   ~  _  2a;  -  4 

^  •  "3"  "^  4a;  -  3       2a;  -  5  ~        3      ' 


57.  a;  +  '^^^^^  —  Viiix"  +  a;''  =  l^w. 


58.    Va;  +  17  +    i/a;  -  4  =  ^  V^ 

r  a;  —  4  oa;  +  1  '  2 

61.  y^^  +  3  +  V3a;-3  =  10.    62.    Vx -{-  V4+^  =  -4^. 
2  2 

63. ; ; =  m. 

x+V-Z  +  x'        X-  V2  +  x' 
,  3 2_ ^ 

^*  a;^  -  7a;  +  3       a;^  +  7a;  +  2 

^'  +  V^  _  a;^  —  a; 

^^    2a;^       .    ,   5a;^       51 

66.  --6+-  =  -^-~^«. 

8a;^  +  10  _   x'  +  4:  __  3^ 
^'         21  5a;^  -  4  ""    8  • 

16 


68.    Vi6  +x  +  Vx  = 
69, 


4/a;-5 
4^  +  3  _    Vx  +  6 


Vx  +  1        Vx  +  2 
70.    ^a;  +  Vx  —  ^x  —  Vx  =  a  y 


x+V 


X 


m  -\-nx       a  +  ex  4  + 3a;         a;  +  2 

71.         J Z=      ; .  72  ~  ' 


73 


mx  -\-  n~  ax  +  c  '  '    S  +  x    ~  2a;  —  19* 

a;''  —  2a;  +  1  __  :r  +  3 
•   x'  +  5a;  +  6  ~  a;  +  2* 


74.    V4a;  +  1  -  i^lO  -a;  =  Vx  +  3. 
12  8  6 


516  qUADUATIG  EQUATIONS. 

2 
77.  6x'  —  7x  =  -  —  4:X.  78.   {x  —  3)  (^  —  4)  =  2. 

81. 

82. 

84. 

86. 

a;  —  1  X  —  6 

1  4a;ir  —  5_^  4;y  +  '^ 

^^*   2x  +  6       3  (:r  +  2)  ~  20'    ^'   T  +  F+1  ""  "^lo"' 

2:^^  —  3    ,    13  _  1-52: 

^°'  4:c  -  1  "*"  ¥  ""  2^  +  3' 


91.  ^x''  —  X-  6  Vx'  +  42^+4  —  42;  +  7, 

92.  x'  -  6x  +  12  +  y:r'  +  6a:  +  9  =  5. 

93.  2a;'  -  2a:  +  6  Vx'  -  x  +  7  =  22. 

94.  3a;  —  2  Va;^-  3a:  +  9  =  26  -  a:'. 


95.  2a:*  +  7a:  -  31  ==  a;  +  4/0;^  +  3a;  +  7. 

96.  y'2a;+  3  -  V a;  -  7  =  3. 
t^2^ T^  -  -/a:*  -  3   _  1^ 

^^'     |/2,,;-  4-  1  +  Vo:^  -  3   ~  ^' 

98.  V2a;  -  3  +  l/4a:  +  1  =  >^/6^+78. 

99.  3  Va;'  +  5  -  V9^~+4^~+~5  ~  2. 

100.  4/0:'  +  3a:  +  /a;'  +  a:  +  2  =  4. 

§  198. 

Factor  the  following  expressions  by  adding  such  a  quan- 
tity as  will  make  the  trinomial  a  perfect  square,  and  subtract- 
ing the  same  quantity. 

a"^  ■—  2ah  —  ZW.     Add  4Z>^  and  subtract  it;  then 

a^  -  2ah  +  Z>'  -  ^y  =  {a-  by  -  W 

=  [{a  -b)-  2b]  [{a  -'b)  +  2h]. 


WITH  TWO    UNKNOWN  QUANTITIES.  517 

'  I.  a'  +  '^a'h'  -  %\  2.  x'  -  2ax  -  3a^ 

3.  x"  -  4cX  +  3.  4.  2:'  +  82:?/  --  9y\ 

5.  a'  -  20abc  +  64^V.  6.   2;^;'  -  ^x  +  3. 

7.  ^'^  +  2%  -  8b\  8.  4^^  -  4.ab  -  Ub\ 

9.  :?;'  +  6x  +  5.  10.   6a'  +  bab  -  6b\ 

II.    -  x'  +  x'  +  12.  12.  a'  +  9«^<^^  +  S\b\ 

13.  2a'  -  2ab  +  b\  14.  a'  +  4.ab  +  3b\ 

15.  x'  -  6x'  -  16.  16.  x'  +  a^y. 

17.  x'  —  Qx'y'  +  Dx'f,  18.   12^'  -f  24a:y  +  %'. 
19.   ia'  -  37 a' b'  +  9>. 

(§3O70        Simultaneous  Quadratics. 

Two  Unknovvk  Quantities. 

I.  X    -\-  y  =    7,  2.  X  —    y    =    5. 

x'  +y'=  25.  x'  -  2xy  =  21. 

3.  22;  —   3?/  =    1.  4.   2a:  —    2?/  =  5. 

3x^—  4cxy  =  15.  5a;'  —  3a:^  —  v''  —  161- 

5.  a;'  +  ^'  =  a'.  6.  a;    +  y  =    28. 

x"  -  y''  =    b\  xy  =  147. 

7.  x'  +  If/'  =  169.  8.  o;^  +  y^'  =:  224. 

^^  =  60.  xy  =  12. 

9.x  +  y=      8.  10.  --  =  --. 

x'+  y'=  224.  3a;y  +  2x  +  y  =  485. 

II.  a;  +  y  +  4/^y  =  19.  12.   lOa;  +  ^  =  3^:?/. 

x'  +  y'  =  97.  .V  =  ^  +  '^^' 

13.  ^'  =  ia;y.  14.  X  +    2y  =:  30. 

cc    -    7/    =  15.  .V'  -  10^  ==  1^^  +  ^6- 

15.   2a;  +  3?/  =  17.  16.   3a;  +  by  =  31. 

xy  =  12.  xy  +  y'=  18. 

17.  5a;  —  3y  =  1. 

2^'  -  a;'  -  3a:f/  +  10a;  -  5^  =  1. 

18.  4a;      —  5^   =  1. 

lly'  -  6x'  -  9xy  +  22a;  -  7y'  =  20. 

19.  7a:^  -  13xy  +  5/  =  -  5. 20.    3a;'  -  lla;y  +  7y'  =  7. 
6x^  -  9xy    +  4:t/  =  6.  5a;'  -  17a^^  +11/  -  1*^- 


518  QUADRATIC  EQUATIONS. 


21. 

S^r''  -  nxy  +  7^/'  =  5. 

22. 

Zx"  -  2^xy  +  ty'  =  15. 

Qx'  -  Ihxy  +  9^/^  =  15. 

7^'  -  48^'^  +19/=  11. 

23- 

X    +  y   =    ij. 

24. 

a;    +  ^   =        8. 

X      +  ;?/    =  72. 

2:^  +  y'  =  3368. 

25- 

X    +  y    =  5. 

26. 

.t'         —  2/'  =  56. 

^'  + .'  =  '7- 

^^  (^  --  ^)  =  16. 

27. 

X     J^  y    ^xy. 

28. 

xy  +  a.y  =  18. 

xy  =  x'  -  y\ 

X    -{-  xy^  =:  27. 

29. 

x'  +y'  +  x  +  y^l^. 

30. 

:z;^   +  ^^     -  a;  -  ?/  =  78. 

xy  =  6. 

^y  +  ^'    +  .V  =  ^^' 

31- 

1  +  i  =  ^. 

^        ^        6 

32. 

3y^  -  2a;'  =  19. 

r?;  +  ^   =  5. 

/  +  a;?/   =  15. 

33- 

x\-\-y\  =z      6. 

34. 

2.r''  +  ?^xy  =  26. 

:rf  +  ^1  .=  126. 

3?/'  +  )lxy  =  39. 

•^ 

36. 

^'^^^    -  'ixy  =  24. 

35- 

"^^y  -x-y 

820 

a;^  4-  ^'^  = . 

a:i/  -  2.?/'  =4. 

37. 

,    a;        5 
^•^  +  .^-3- 

38. 

^  +  .y  =  ^+2. 

1  4.  ^  -  io 

~T~           "~~      0  • 

^  +  ^  =  ^5: 

xy       X         3 

y     ^ 

39- 

^x'  +  3/  =  27. 

40. 

4.x'  -  57/'  =  16. 

^x"  -  .^'^    =  15. 

^x'  +  2^^^  =  35. 

41. 

x'    -^y'    =45. 

42. 

Zx  -  2y  =  6. 

a:     =2y. 

:r?/  ~  .1;     =8. 

43- 

xy  =12. 

44. 

4cX^  —  5y  =  4:xy. 

3a;  —  2.y  =  1. 

5x  +  3y  =  37. 

45. 

X     -\-  xy  =  24. 

46. 

^'  +  ^'  +  ^  +  ^  =-  36. 

y      4-  2:?/    =  21. 

^^  -  2/'  +  ^  -  2/  =  ^4. 

47. 

x'  +  xy=zd5. 

48. 

x'  +  xy  +  f  =  7. 

y''  -f  o;^^  =  14. 

x'-xy  +  y'  =  3. 

49. 

x'  ~  2i:«/  +  ^'    =7. 

50. 

^  +  ^y  +  y  =  11- 

x'  -  Zxy  +  2^^  =  -  2. 

^'  +  ^y  +  y"  =  19. 

WITH  THREE  UNKNOWN  QUANTITIES.  gig 

52.  x""  +  y  =.  ^{x-y). 

X  +y=4(a;-,v). 

^y.  54.  x^  —  xy^^l^, 

x^  J^y'  =  74. 
56.  x''  -  ^xy  +  2y^  =  1. 
x"^  +  2.T^  —  4y'^  =  5. 
58.       i^+     Vy  =z    5. 

ic  V.'c  -f-  ^  '^.V  =  35. 
60.  4:x'  -  9y'  =  7. 

2x  +^  =7. 
62.  a:'  +  y^  =  25. 

X  +y   =    7. 
64.  X  —  2y  =      2. 

x'  +  4y'  =  100. 

Three  or  more  Unk:n^ovvn  Quantities. 

I.  xy  =  24.  2.  x""  +  .T2;  =  24. 

{7  -y)z  =  8.  ,-z'  +  XZ  =  12. 

(3  -x)  {z-  11)  =  3.  y'  +  yz  +  z'  =  28  -  3?/. 

3.  :/;?/2;  =  3  (.t^  +  4)  =  12  (2;  +  ^)  =  4  (o;^  +  ;^  -  10). 

4.  :z:'  +  .y'  +  ^'  =  84.  5.   ^  +  ^Z  +  ^  =  14. 

•^  +  .V  +  ^  =  14.  a:'  +  y'  +  z'  =  84. 

:r7/  =  8.  xz  =  y\ 

X  -\-  y  _b 
6.  a:  +  f/  +  2;  =r  12.  7-   — ^  -  g . 


5^. 

x  +  2  _  ^ 
x'+  2       19 

53. 

55. 

^+2"   3* 
2-{-y    =  40  —  X 
:.  +  2^  =-  7. 
x^  —  xy  =  35. 
^^_  ^^  =  10. 

57. 

.T^'  +  ^:y  =  18. 

2;?/^  +  a;    =  27. 

59- 

X  -y   =    8. 
rr^  -  ?/^  =  80. 

61. 

4  (.r  +  2^)  =  12. 
rr'^      -  4//*^  :=  33. 

(>z- 

X   —  y    =    2. 
ic^  4-  2/^  =  34. 

^y  -\-  yz  -}-  zx  =  47. 


a;+  ^       3 


xz  4' 

^•'  +  2/'  -  ^'  =  0.  y  +  ^  ^  _7 

y^  12* 

8.   2x'  +  2.T^y  +  y'  =  49.  9.  :x^  -  ^^y  +  2^  =  2. 

x'  —  .T2;  +  ;a^'  =  28.  x'  +  f  +  /  =  49. 

//  +  2yz  +  i^'  =  25.  xy  —  z-\-y  —  3. 

:^  +  ?/  +  2;  =  9.  II.  a;  +  ^'zy  +  2;  =  10. 

"  +  y/'^  +  ;2;^  =  29.  x^  +'^«  +  ^'  =-  38. 

^'  =  4^  +  1.  ^^  _|_  ^2;  =  .t'. 


10. 
.7; 


520  QUADRATIC  EQUATIONS, 

12.  x-{-  y  =1,  1^,  X  -\-  y  =  9. 

II  J^  V  =  1.  u  -\-  V  =  ^, 

X  +  u'  ^  S,  x'  +  u'  =  52. 

y  +  v'  :=  4.  y'  +  v'  =  41. 

14.  xii  =  yv.  15.  xy  =  35. 

a;  -[-  y  =14.  tcv  =  18. 

^^4-^=7.  x  +  u  =  13. 

~+^=    4.  ^  +  i;  =9. 

Problems  Leading  to  Quadratic  Equations. 

1.  A  principal  of  $6000  amounts  with  simple  interest  to 
17800  after  a  certain  number  of  years.  Had  the  rate  been  1 
per  cent,  higher  and  the  time  1  year  longer,  it  would  have 
amounted  to  $720  more.     What  was  the  time  and  rate? 

2.  A  courier  left  a  town  riding  at  a  uniform  rate.  Three 
hours  afterwards  another  followed,  going  1  mile  an  hour 
faster.  Two  hours  after  the  second  another  started,  going  6 
miles  an  hour.  They  arrive  at  their  destination  at  the  same 
time.     What  was  the  distance  and  rate  of  riding? 

Ans.  Dist.  =  60  or  6.     Speeds,  4,  5  and  6  or  1,  2  and  6. 

3.  In  aright-angled  triangle  the  hypothenuse  is  5  and  tlie 
area  6.     What  are  the  sides? 

4.  Find  two  numbers  whose  product  is  180,  and  if  the 
greater  be  diminished  by  5  and  the  less  increased  by  3,  the 
product  of  the  sum  and  difference  will  be  150. 

5.  Find  two  numbers  whose  sum  is  100  and  the  sum  of 
their  square  roots  14. 

6.  Find  two  numbers  whose  sum  is  35  and  the  sum  of 
their  cube  roots  5. 

7.  By  selling  a  horse  for  $130  I  gain  as  much  per  cent,  as 
the  horse  cost  me.     What  did  I  pay  for  him? 

8.  What  is  the  price  of  apples  a  dozen  when  four  less  in 
20  cents'  worth  raises  the  price  5  cents  per  dozen? 

9.  The  sum  of  the  squares  of  three  consecutive  numbers  is 
149.     What  are  the  numbers? 

10.  If  twice  the  product  of  two  consecutive  numbers  be 
divided  by  three  times  their  sum  the  quotient  will  be  f .  What 
are  the  numbers? 


PROBLEMS.  521 

11.  A  womnn  bought  a  number  of  oranges  for  3G  cents. 
If  she  had  bought  4  more  for  the  same  money  she  would  have 
paid  i  of  a  cent  less  for  each  orange.     How  many  did  she  buy? 

12.  In  mowing  60  acres  of  grass,  5  days  less  would  have 
been  sufficient  if  2  acres  more  a  day  had  been  mown.  How 
many  acres  were  mown  per  day? 

13.  A  broker  bought  a  certain  number  of  shares  (par  $100 
each)  at  a  discount  for  $6400.  When  they  were  at  the  same 
per  cent,  premium,  he  sold  all  but  20  for  $7200.  How  many 
shares  did  he  buy,  and  at  what  price? 

14.  If  the  length  and  breadth  of  a  rectangle  were  each  in- 
creased by  2,  the  area  would  be  238;  if  both  were  each  dimin- 
ished by  2,  the  area  would  be  130.  Find  the  length  and 
breadth. 

15.  Twice  the  product  of  two  digits  is  equal  to  the  number 
itself;  and  7  times  the  sum  of  the  digits  is  equal  to  the  number 
formed  by  the  same  digits  reversed.     What  is  the  number? 

16.  The  sum  of  two  numbers  is  ^  of  the  greater,  and  the 
difference  of  their  squares  is  45.     What  are  the  numbers? 

17.  The  numerator  and  denominator  of  two  fractions  are 
each  greater  by  2  than  those  of  another,  and  the  sum  of  the 
two  fractions  is  2-|;  if  the  denominators  were  interchanged, 
tlie  sum  of  the  two  fractions  would  be  3.  What  arc  the  frac- 
tions? 

18.  A  man  starts  from  A  to  go  to  B.  During  the  first  half 
of  the  journey  he  drives  \  mile  an  hour  faster  than  the  other 
half,  and  arrives  in  5f  hours.  On  his  return  he  travels  a  mile 
slower  during  the  first  half  than  when  he  went  in  going  over 
the  same  portion,  and  returned  in  6f  hours.  What  was  the 
distance  and  rate  of  driving? 

19.  .A  person  who  has  $8800  invests  a  part  of  it  in  one 
enterprise  and  the  rest  in  another;  the  dividends  differ  in  rate, 
but  are  equal  in  amount.  If  the  sums  invested  liad  exchanged 
rates  of  dividends,  the  first  would  have  yielded  $200  and  the 
other  $288.     What  were  the  rates? 

20.  Divide  50  into  two  such  parts  that  their  product  may 
be  to  the  sum  of  their  squares  as  6  to  13. 

21.  A  company  at  a  hotel  had  $12  to  pa}^  but  before  set- 


522  QUADRATIC  EQUATIONS. 

tling  2  left,  when  those  remaiiiiiig  had  30  cents  apiece  more 
to  pay  than  before.     How  many  were  there? 

2  2.  A  drover  bought  a  nnmber  of  sheep  for  $180;  after 
keeping  10  he  sold  the  rest  for  $200,  and  gained  33^  cents 
apiece.     How  many  did  he  buy? 

23.  Two  partners,  A  and  B,  gained  $140  in  speculation; 
A's  money  was  3  months  in  trade,  and  his  gain  was  $60  less 
than  his  capital;  B's  money,  which  was  $50  more  than  A's, 
was  in  5  months.     What  was  each  man's  capital? 

24.  Divide  30  into  two  such  parts  that  their  product  may 
be  36  times  tlieir  difference. 

25.  A  aiid  B  set  out  from  two  towns  which  are  126  miles 
apart,  and  travelled  until  they  met.  A  went  8  miles  an  hour, 
and  the  number  of  hours  they  travelled  was  3  times  greater 
than  the  number  of  miles  B  travelled  an  hour.  What  were 
their  hourly  rates?  A^is.y  in  part.   B's  rate,  VoS  —  4. 

26.  In  a  purse  containing  28  pieces  of  silver  and  nickel, 
each  silver  coin  is  worth  as  many  cents  as  there  are  nickel 
coins,  each  nickel  is  worth  as  many  cents  as  there  are  silver 
coins,  and  the  whole  are  worth  $1.50.  How  many  are  there 
of  each? 

27.  Find  two  such  numbers  that  the  product  of  their  sum 
and  difference  may  be  7,  and  the  product  of  the  sum  and  dif- 
ference of  their  squares  may  be  144. 

28.  A  grocer  received  an  order  for  12  pounds  of  sugar  at 
12  cents  a  pound.  If  he  should  have  none  for  that  price,  he 
was  to  send  as  many  pounds  more  or  less  than  12  as  the  sugar 
cost  less  or  more  than  12  cents  a  pound.  The  bill  amounted 
to  $1.35.  How  many  pounds  had  he  sent,  and  what  was  the 
price  per  ]>ound? 

29.  A  grocer  sold  50  pounds  of  pepper  and  80  pounds  of 
ginger  for  $26;  but  he  sold  25  ])ounds  more  of  pepper  for  $10 
than  he  did  of  ginger  for  $4.  What  was  the  price  per  pound 
of  each? 

30.  A  and  B's  shares  in  speculations  together  amounted  to 
$675.  A  had  his  money  invested  5  months  and  B  4^  months, 
and  each  receives  in  capital  and  profits  $455.  What  did  each 
be2:in  with? 


PROBLEMS.  523 

^  31.  A  person  rents  a  certain  number  of  acres  of  land  for 
$120;  he  retains  10  acres,  and  sublets  the  rest  at  20  cents  an 
acre  more  than  he  gave,  and  receives  $12  more  than  he  pays 
for  the  whole.  How  many  acres  were  there,  and  how  much 
per  acre? 

32.  A  person  bought  a  certain  number  of  shares  for  as 
many  dollars  per  share  as  the  number  he  buys;  after  they  rose 
as  many  cents  per  share  as  he  had  shares,  he  sold  them  and 
gained  $4.     How  many  shares  did  he  buy? 

2,2,.  The  income  of  a  certain  railway  company  would  justify 
a  dividend  of  5  per  cent,  of  the  whole  stock;  but  as  $150,000 
of  the  stock  is  prefei-red^  guaranteeing  6  per  cent.,  the  divi- 
dend for  the  remaining  stock  is  reduced  to  4f  per  cent. 
What  is  the  whole  amount  of  stock? 

34.  The  length  of  a  rectangular  farm  is  to  its  width  as  4 
to  3;  f  is  in  grass,  and  the  remaining  45  acres  is  cultivated. 
What  are  the  dimensions  of  the  field? 

35.  If  a  straight  line  be  divided  into  two  such  parts  that 
the  rectangle  contained  by  the  whole  line  and  one  part  is  equal 
to  6  times  the  square  of  the  other  part,  what  will  be  the  ratio 
of  these  two  parts? 

2,6,  Out  of  a  sphere  of  clay  whose  diameter  is  16  inches, 
two  spheres  are  formed  with  radii  of  3  and  5  inches  respec- 
tively. If  the  volumes  of  spheres  vary  as  the  cubes  of  their 
radii,  what  will  be  the  radius  of  the  sphere  that  can  be  made 
of  the  clay  that  remains? 

37.  The  two  digits  of  a  certain  number  differ  by  1,  and 
their  product  is  i  of  the  noxt  higher  number,  what  is  the 
number? 

38.  Find  five  numbers  having  equal  differences,  and  such 
that  their  sum  shall  be  40,  and  the  sum  of  their  cubes  3520. 

39.  A  merchant  bought  a  barrel  of  wine  for  $60;  he  re- 
tained 12  gallons  for  his  own  use  and  sold  the  remainder  at 
an  advance  of  80  per  cent,  on  each  gallon  and  gained  20  per 
cent,  on  the  whole.     At  what  price  per  gallon  did  he  sell  it? 

40.  Find  two  numbers  that  are  to  each  other  as  9  to  7; 
and  the  square  of  their  sum  is  equal  to  the  cube  of  their  dif- 
ference. 


524  QUADRATIC  EQUATIONS. 

41.  The  panel  in  a  door  is  12  by  18  inches,  and  it  is  to  be 
SRiTouiided  by  a  margin  of  uniform  width  and  equal  surface 
to  the  panel.     How  wide  must  the  margin  be? 

42.  The  fore  wheel  of  a  coach  makes  6  more  revolutions 
than  the  hind  wheel  in  going  160  yards;  but  if  the  circumfer- 
ence of  each  wheel  be  increased  by  4  feet,  the  fore  wheel  will 
make  only  4  more  revolutions  in  160  yards.  What  is  the  cir- 
cumference of  eacli  wheel? 

43.  The  sum  of  three  numbers  is  15;  the  difference  between 
tlic  first  and  third  is  3  more  than  the  difference  between  the 
second  and  third,  and  the  sum  of  their  squares  is  93.  What 
are  the  numbers? 

44.  Tiie  product  of  two  numbers  is  15,  and  if  their  differ- 
ence be  added  to  the  difference  of  their  squares  the  sum  will 
be  18.     What  are  the  numbers? 

45.  A  certain  number  consists  of  two  digits;  the  number 
is  4  times  the  sum  of  its  digits;  and  3  times  the  number  is 
equal  to  twice  the  square  of  the  sum  of  its  digits.  What  is 
the  number? 

46.  Find  two  numbers  whose  sum  is  14,  and  if  their  prod- 
uct be  added  to  the  sum  of  their  squares  the  result  will  be 
148. 

47.  Two  brokci's  begin  business  with  a  joint  capital  of 
$10,000.  A  withdraws  at  the  end  of  12  months  and  receives 
$4960  in  capital  and  profits.  B  remains  3  months  longer  and 
receives  $7800  stock  and  gain.  What  was  the  original  capital 
of  each? 

48.  Find  five  equal  numbers  whose  sum  is  equal  to  their 
continued  product. 

49.  A  jockey  bought  a  horse  and  sold  it  at  a  certain  per 
cent,  profit;  with  the  money  he  bought  another  horse  and 
sold  it  at  the  same  per  cent,  profit,  and  with  the  proceeds  he 
was  able  to  buy  2  horses  each  costing  2  per  cent,  less  than  the 
first.     What  per  cent,  did  he  make  on  each  transaction? 

50.  Two  travellers  start  from  the  same  place  at  the  same 
time,  one  goes  due  north  16  miles  a  day,  and  the  other  due 
east  21^  miles  a  day.  How  long  must  they  travel  in  order  to 
be  160  miles  apart? 


PROBLEMS.  525 

51.  What  is  the  length  of  a  side  of  a  square  whose  area  is 
increased  by  |of  its  amount  when  4  feet  is  added  to  each  side? 

52.  Find  the  length  of  the  side  of  a  square  such  that  the 
number  of  square  feet  in  its  area  exceeds  the  number  of  linear 
feet  in  its  perimeter  by  12. 

53.  The  perimeter  of  a  rectangle  is  34  feet;  if  its  length 
were  increased  by  4  feet,  while  its  perimeter  remained  the 
same,  the  former  area  would  exceed  the  double  of  the  second 
by  6  feet.     What  were  the  original  dimensions? 

54.  If  3  feet  be  taken  from  one  side  of  a  rectangle  whose 
perimeter  is  14  feet  and  added  to  the  other  side,  the  area 
would  be  doubled.     What  were  the  first  dimensions? 

55.  A  man  invests  his  money  at  a  certain  rate  of  interest 
for  two  years,  and  finds  that  he  will  get  1  per  cent,  more  for 
it  if  he  reckon  by  compound  interest  compounded  annually 
than  by  simple  interest.     What  is  the  rate  of  interest? 

56.  A  person  bought  a  certain  number  of  shares  when  they 
were  at  a  discount  and  sold  them  when  they  rose  to  a  premium 
of  the  same  rate  per  cent.  His  profit  on  the  first  investment 
was  A  percent,  more  than  the  common  value  of  the  premium 
and  discount.     What  was  the  latter  and  the  rate  of  profit? 

57.  A  regiment  of  2196  soldiers  is  formed  into  two  squares, 
one  having  6  more  men  on  a  side  than  the  other.  How  many 
men  are  there  on  a  side  of  each  square? 

58.  Find  two  numbers  whose  product  is  twice  their  sum, 
and  the  sum  of  their  squares  45. 

59.  Find  two  numbers  whose  product  is  8  times  their  dif- 
ference, and  the  difference  of  their  squares  48. 

60.  Find  two  numbers  whose  difference  is  6,  and  |  of  their 
product  is  equal  to  the  square  of  the  less. 

61.  Find  two  numbers  such  that  their  product  added  to 
twice  the  square  of  the  greater  is  65,  and  the  product  added 
to  the  square  of  the  less  is  24. 

62.  Find  two  numbers  such  that  their  sum  multiplied  by 
the  sum  of  their  squares  is  715,  and  the  difference  multiplied 
by  the  difference  of  their  squares  is  99. 

6^,  Two  trains  start  at  the  same  time  from  two  towns  and 
run  at  a  uniform  rate  towards  the  other  town.     When  Ihcy 


526  PROGRESSIONS. 

meet  it  is  found  that  one  train  has  travelled  90  miles  more 
than  the  other,  and  that  if  they  continue  at  the  same  rates 
they  will  finish  the  journey  in  6  and  13|-  hours.  What  are 
the  distance  and  rates? 

64.  A  man  receives  $2200  a  year  interest.  If  he  had  in- 
vested his  capital  at  ^  per  cent  higher,  he  could  have  lessened 
his  investment  by  $4000  and  received  the  same  income  as  be- 
fore.    How  much  had  he  invested? 

Progressions. 

Note.— Tlie  abbreviations  A.  P.,  G.  P.,  C.  D.,  and  C.  R.  are  but  for 
Arithmetical  Progression,  Geometrical  Progression,  Common  Difference, 
and  Common  Ratio,  respectively. 

1.  If  the  first  and  last  terms  of  an  arithmetical  progression 
are  a  and  I  and  the  number  of  terms  n,  express  the  sum  of 
all  the  intermediate  terms. 

2.  If  the  first  and  last  terms  of  an  A.  P.  are  4  and  28 
respectively,  what  possible  values  may  the  sum  of  the  inter- 
mediate take? 

3.  Sum  to  n  terms  distinguishing  the  cases  when  n  is  even 
and  odd,  when  necessary: 

1-3+5-7+ 

4.  2-4+6-8+ ■ 

5.  p,p  +  n,p+2n, 

6.  If  the  square  of  the  fourth  term  of  an  A.  P.  is  equal  to 
the  product  of  the  first  and  sixth,  show  that  the  tenth  term 
must  vanish. 

7.  If  the  square  of  the  second  term  of  an  A.  P.  is  equal  to 
the  product  of  the  first  and  fourth,  show  that  tlie  square  of 
the  sixth  is  equal  to  the  product  of  the  fourth  and  ninth. 

8.  Generalize  the  preceding  result  by  showing  that,  in 
order  that  the  square  of  tlie  nth.  term  may  be  equal  to  the 
product  of  the  first  and  n'ih,  and  the  square  of  the  mth  to 
the  product  of  the  ^^'th  and  m'th,  it  is  necessary  and  sufficient 
that  m,  m',  n  and  n^  fulfil  the  conditions 

m'  =  2  {m  —  7i)    +1;     2u^  =  ni  +  n 

9.  Find  three  quantities  in  A.  P.  whose  sum  shall  be  3« 
and  the  sum  of  whose  squares  shall  be  ll«^ 


PUOGRESSIONS,  527 

10.  Find  7  terms  of  an  A.  P.  such  that  their  sum  shall 
be  14  and  the  sum  of  their  squares  84. 

11.  In  an  A.  P.  the  product  of  the  first  and  eighth  terms  is 
less  by  h  than  the  product  of  the  second  and  seventh.  How 
much  less  is  the  product  of  the  third  and  sixth  than  that  of 
the  fourth  and  fifth? 

12.  Express  the  sum  of  n  terms  of  an  A.  P.  in  terms  of 
the  first  term  and  the  CD. 

13.  If  a  and  h  are  the  first  two  terms  of  an  A.  P.,  express 
the  last  term  and  the  sum  of  n  terms. 

14.  Prove  that  if  the  sum  of  m  terms  of  an  A.  P.  be  n, 
and  the  sum  of  n  terms  be  m,  we  shall  have 

2  (m  +  ^0  +  ^^^  —  ^• 

15.  If  a^,  If,  &  be  ill  A.  P.,  then, 


a-\-V     c  -\-a'     h  -\-  c 

will  also  be  in  A.  P. 

16.  The  sum  of  the  first  three  terms  of  an  A.  P.  is  15  and 
the  sum  of  their  squares  is  83.     What  is  tlie  sum  of  n  terms? 

17.  In  a  progression  of  9  terms,  the  third  term  is  10  and 
the  sum  153.     Find  the  first  term  and  common  difference. 

18.  In  an  A.  P.  a  certain  term  is  h\  there  are  2n  terms 
before  k  and  n  terms  after  it,  and  the  sum  of  all  the  terms  is 
3^  +  1.     Find  the  0.  D. 

19.  Two  men  start  simultaneously  from  the  same  point  in 
the  same  direction.  The  one  walks  m  miles  the  first  day, 
and  diminishes  his  walk  by  h  miles  each  day;  the  other  walks 
n  miles  the  first  day,  and  increases  his  walk  h  miles  each  day. 
How  far  will  the  latter  be  ahead  at  the  end  of  i  days? 

20.  Express  the  sum  of  the  G.  P.'s: 

.    ^n  _(_  ^2n  _|.  ^3n  _|_   _    _    _^  ^lOn. 
^n  _(_  ^2n  ^  ^3n  _|_   _    _    _|_  ^mn.^ 

1     +  |/3  +  3      + +  3^ 

21.  The  sum  of  the  first  and  seventh  terms  of  a  G.  P.  is 
A,  and  the  sum  of  the  second  and  eighth  is  h.  Find  the  first 
term  and  the  C.  K. 

22.  The  sum  of  the  first  and  fifth  terms  of  a  G.  P.  being 


528  PROGRESSIONS. 

added  to  twice  the  third  term  gives  a  sum  which  is  9  times 
the  first  term.     Find  the  C.  R. 

23.  The  fifth  term  of  a  G.  P.  exceeds  the  first  by  16,  and 
the  fourth  exceeds  the  second  by  4  1^3.  Find  the  first  term 
and  0.  R. 

24.  In  a  G.  P.  the  sum  of  n  terms  is  S  and  the  sv>  WjI  2n 
terms  is  6S,     Express  the  0.  R.  and  first  term. 

25.  In  a  G.  P.  of  271  -f  1  terms,  whose  first  term  is  5,  the 
sum  of  the  first  and  last  terms  is  125  greater  than  twice  (he 
middle  term.     Find  the  C.  R. 

26.  The  first  term  of  a  G.  P.  is  2,  and  the  continued 
product  of  the  first  5  terms  is  128.     What  is  the  0.  R.  ? 

27.  Find  that  G.  P.  of  which  the  product  of  the  first  and 
second  terms  is  3,  and  that  of  the  third  and  fourth  terms  is  48. 

28.  A  person  who  each  year  gained  half  as  much  again  as 
he  did  the  year  before,  gained  $2059  in  7  years.  What  was 
his  gain  the  first  year? 

29.  A  man  who  had  a  principal  out  at  5  per  cent,  per 
annum  compound  interest  for  4  years  found  that  the  interest 
gained  during  the  second  and  fourth  years  was  greater  by 
$84.10  than  that  gained  during  the  first  and  third  years. 
What  was  the  principal? 

30.  Show  that  \ia,h,c,d,  ,  .  .  ^,  ?  be  in  G.  P.  we  shall  have 
(a  +  ^  +  c  + -{-k)  {h  -^  c^  d-\- +  /) 

=  ^(^  +  ^  +  ^  +  ....  +  0^ 

31.  If  a,  h,  c,  d  be  in  G.  P.  prove  that 

(^«  J^V"  -\-  c')  {V  +  c'  +  d')  =  [ah  +  ic  +  cd)\ 
{h  -  cy  +  (c  -  ay  -\-  {d-  by  =:  (a-  dy. 

32.  Generalize  the  first  of  the  preceding  results  by  show- 
ing that  if  we  multiply  the  sum  of  the  squares  of  the  first  n 
terms  of  a  G.  P.  by  the  sum  of  the  squares  of  the  n  terms 
following  the  first  term,  the  product  will  be  equal  to  the 
square  of  the  sum  of  all  the  products  formed  by  multiplying 
each  term  from  the  first  to  the  ^th  by  the  term  following  it. 

33.  Sum  to  71  terms 

(^_Ly+(,«»_i,y+(,«^-i,)V.... 


PIWGRESSI0N8.  629 

34-   I^^  *i  Cr-  P-  <^f  ^  terms  are  given: 

The  sum  of  all  the  terms  except  the  first  =  33; 
The  sum  of  all  the  terms  except  the  last  =  —  22. 
Find  the  series. 

35.  Find  two  quantities  of  which  the  arithmetical  mean  is 
a  and  th.    geometrical  mean  is  g,  and  prove  the  result. 

2,6.  In  a  G.  P.  of  8  terms  the  product  of  the  four  alternate 
terms  beginning  with  the  first  is  1,  and  the  product  of  tlie 
four  alternate  terms  from  the  second  to  the  eighth  is  16. 
Find  the  progression. 

37.  A  party  of  m  persons  have  s  dollars  unequally  divided 
among  them.  Each  simultaneously  divides  his  money  equally 
among  his  m  —  1  fellows.  If  one  of  the  party  had  a  dollars 
in  the  beginning,  how  much  will  he  have  after  1.2,  and  p 
such  divisions? 

'    7)1      m  —  l\m        J      m       (m  — l)\m        / 

m       {m  —  1)^  \m         ) 
Find  the  limits  of  the  sums  of  the  progressions: 

38.1  +  1  +  1+.... 

39.  -  +  !  +  «+.... 
n  m 

7171^' 

43..1  +  (r  +  i)"'+(r+3-)"%.... 

43.  l-(r  +  9"V(r  +  i 

44.  r+{l+ay+{l+a+a'y+{l+a+a''+a'y+ 

r  and  ar  being  each  less  than  unity. 

45.  r+{l-ay+{l-a+a'y+{l~a+a''~ay+ 


530  FUNCTIONAL  NOTATION 

46.  r-{l-aY+(l-a+a'y-{l-a+a'-a'y-\-  .  X[. 

^'''  n  +  l       {n+  ly  "^  (^n  +  ly      

Functional  Notation. 

Prove: 

1.  {2n)l  =  2^{l.d.5.  .  .  .  2/^-l)  .nl 

2.  (2^)  !  =  2^^  (1.3.5 15)  (1.3.5.?)  (1.3). 

Using  the  notation  [m]  =  1.3.5.7  .  .  .  .  m 

y^  =  2« 
Show  that  we  have 

3.  ^!  =  2''-[^-l][|-l][|-l].  ...  [3]. 


[2u  -  ly  ,j 

5.  If  S{)i)  represent  the  sum  of  the  first  n  terms  of  a -- 
geometrical  progression  whose  0.  R.  is  r,  show  that 

S{2n)  =  (r^  +  l)S{?i). 

6.  What  will  be  the  last  factors  in  the  numerators  and 
denominators  of  the  following  expressions: 

m-  i"^)--  ^"^y-  (^D^  c^i)^  (^')- 

«-e)+c-ii)=(:^>. 


FUNCTIONAL  NOTATION.  531 

Sn  represent  the  sum  of  the  first  n  natural  numbers, 
^hat    s, 

/y,=  l  +  2  +  3  + +n, 

aow  that: 

10.  /S'n  :    Sn  +  1  =  n  :  n  +  2. 

11.  ^2X  /S',  X  /5i  =  3!  [7]. 

12.  .^2  X  ^S',  X  >S; X  >^2n  =  nl  [2n  +  1]. 

13.  S,XS,XS, X  .%n+i  -  (^  +  1)!  [2n  +  1]. 

14.  S,X  S,X  S, X  S,n  =  {2n  +  1)  (t^!)^  [2^  -  l]^ 

15.     /S'i+.?,+  ^8+ +^2. 

=  4.{V  +  2'  +  d'  +  4:'+ +  7i'). 

6.     S,+  S,+  S,+ +  S,r^  +  ^ 

==  1^  4_  3«  +  5«  +  .  .  .  .  4.  (27^  +  ly. 

17.  IlCi=^h  +  sCi^i  find  the  values  of  0^,  C3,  Ci,  and  Of 
terms  of  A,  5  and  Co,  and  find  the  value  toward  which  6^ 

oaches  as  i  increases  indefinitely,  assuming  5  <  1. 

1 8.  Apply  the  preceding  notation  to  the  following  problem : 
A.  person  having  a  full  and  an  empty  cask  pours  half  the  con- 
tents of  the  full  one  into  the  other;  then  half  of  this  last  one 
back  again.  He  repeats  this  double  operation  an  indefinite 
number  of  times.  Find  what  fraction  of  the  liquid  will  re- 
main in  the  first  cask  after  1,  2,  3,  4,  and  i  such  double  opera- 
tions. 

To  do  this  assume  that  d  and  1  —  d  represent  the  fractions  of  the 
.  aid  in  the  two  casks  after  the  ^th  operation,  and  then  find  the  fractions 
after  the  {t  +  l)st  operation. 

19.  A  vintner  has  one  cask  containing  a  gallons  of  wine 
and  another  containing  b  gallons  of  water.  He  pours  half  the 
wine  into  the  water,  then  half  that  mixture  back  into  the 
wine,  and  so  on  indefinitely.  Find  an  expression-  for  the 
quantities  and  proportions  of  wine  and  water  in  each  cask 
after  2n  and  also  after  2n  -\-l  such  operations. 


532  PERMUTATIONS  AND   COMBINATIONS. 


Permutations  and  Combinations. 

1.  A  regular  cube  is  to  have  its  sides  numbered  1,  2 ....  G. 
In  how  many  ways  may  the  numbering  be  done? 

2.  In  how  many  ways  might  the  numbering  be  done  in 
the  last  problem  if  only  three  of  the  six  sides  were  to  be 
numbered  ? 

3.  A  party  of  3  boys  and  4  girls  has  to  walk  in  single  file, 
the  boys  ahead.     In  how  many  ways  can  they  be  arranged? 

4.  What  would  be  the  number  of  aiTangements  in  the  last 
problem  if  the  only  condition  were  that  the  boys  must  be  to- 
gether in  one  group  and  the  girls  in  another? 

5.  If  the  combination  of  auy  three  different  letters  in  any 
,  order  made  a  word,  how  many  words  of  three  letters  could  be 

formed  from  the  26  letters  of  the  alphabet? 

6.  If  in  the  last  problem  the  words  thus  formed  were 
divided  into  sets  such  that  the  different  words  of  a  set  should 
be  formed  of  the  same  letters,  how  many  sets  would  there  be, 
and  how  many  letters  in  a  set? 

7.  Six  men  with  their  wives  are  to  stand  in  a  row.  In 
how  many  ways  may  they  be  arranged  subject  to  the  condition 
that  each  man  must  remain  alongside  his  wife? 

8.  What  would  be  the  answer  to  the  last  problem  in  case 
each  man  had  to  keep  his  wife  on  his  right? 

9.  A  boy  has  the  letter  blocks  which  form  the  words  you 
are  mad.  In  how  many  of  the  arrangements  will  all  three 
words  be  recognized,  supposing  that  any  word  may  be  recog, 
nizedwhen  its  first  letter  stands  first,  and  its  other  letters  fol- 
low it  in  any  order? 

10.  If  every  permutation  of  two  or  more  letters  made  a 
word,  how  many  words  could  be  formed  from  10  letters? 

11.  In  how  many  permutations  of  n  letters  will  the  first 
letter  retain  its  place?  The  second  letters  retain  their  second 
places?    The  last  letter  retain  the  last  place? 

12.  If  we  write  under  each  other  all  possible  permutations 
of  the  first  n  numbers  1,  2,  3  ....  ^^  what  will  be  the  sum 
of  each  column?  Ans,  ^{n  -\-l)\ 


)  PERMUTATIONS  AND   COMBINATIONS.  533 

13.  What  will  be  the  sum  of  each  column  if  the  possible 
permutations  of  the  figures  1  2  2  3  3  3  4  are  all  written  under 
each  other? 

14.  From  a  collection  of  5  capital  letters  and  7  small  ones 
how  many  combinations  of  1  capital  with  2  small  ones  can  be 
formed? 

15.  The  driver  of  a  four-horse  coach  can  choose  his  horses 
from  a  stable  of  6  white  and  8  black  horses,  but  he  must  not 
pair  2  horses  of  different  colors.  In  how  many  different  ways 
may  he  choose  his  4  horses? 

16.  How  many  of  the  possible  combinations  of  3  letters 
in  the  first  10  will  contain  the  letter  c?  How  many  will  con- 
tain both  the  letters  c  and  e?? 

17.  Of  the  possible  combinations  of  s  things  in  n,  how 
many  will  contain  a  designated  thing?  How  many  2  desig- 
nated things?    How  many  h  designated  things? 

18.  A  party  of  6  meet  for  whist,  2  waiting  while  the  other 
4  play.  Each  4  must  play  one  game  with  each  possible  ar- 
rangement of  partners.  How  many  games  will  be  played  in 
all;  how  many  will  each  person  play,  and  how  many  times 
will  any  two  designated  persons  have  met  as  partners? 

19.  From  a  collection  of  5  letters  and  6  numbers  how 
many  combinations,  each  consisting  of  1  letter  and  2  num- 
bers, can  be  formed?  How  many  consisting  of  2  letters  and 
3  numbers?     Of  5  letters  and  4  numbers? 

20.  From  a  collection  of  m  letters  and  n  numbers  how 
many  combinations  of  r  letters  with  s  numbers  can  be 
formed? 

21.  In  how  many  ways  may  a  pile  of  20  balls  be  divided 
into  two  piles,  the  one  having  15  balls  and  the  other  5  ? 

22.  How  many  different  signals  may  be  made  with  4  flags 
of  different  colors,  it  being  assumed  that  each  different  order 
of  each  combination  forms  a  different  signal,  but  that  the 
signal  remains  the  same  when  the  order  is  reversed? 

23.  What  would  be  the  answer  to  the  preceding  problem 
if  each  combination  of  several  flags  could  be  arranged  either 
horizontally  or  vertically,  and  an  inversion  of  each  vertical 
arrangement  made  a  different  sig,nal? 


634  PERMUTATIONS  AND   COMBINATIONS. 

24.  How  many  different  signals  can  be  made  with  10  flags, 
of  which  2  are  white,  3  red,  and  5  blue,  all  hoisted  together 
in  a  vertical  row? 

25.  How  many  different  arrangements  can  be  made  of  a 
base-ball  "  nine,"  supposing  that  only  one  man  can  pitch,  and 
only  two  can  catch  ? 

26.  Supposing  that,  in  a  game  of  chess,  the  first  player 
always  has  a  choice  of  two  good  moves  and  the  second  player 
of  three,  how  many  games  of  20  moves  each  are  possible? 

27.  If  the  8  pieces  at  chess  could  be  arranged  in  any  order 
on  the  8  squares  of  the  first  rank,  how  many  different  arrange- 
ments would  be  possible? 

28.  In  how  many  different  ways  can  4  pawns  be  arranged 
upon  the  64  squares  of  a  chess-board?  How  many  different 
arrangements  can  be  made  with  a  king,  queen,  knight,  and 
rook?    Explain  the  relation  of  the  two  answers. 

29.  In  how  many  ways  may  12  balls  be  divided  into  three 
piles,  containing,  the  one  3  balls,  the  second  4,  and  the  thirds? 

30.  In  how  many  ways  may  n  balls  be  divided  into  3  piles, 
containing,  the  one  p,  the  second  q,  and  the  third  r  balls 
{p+q  +  r^^n)? 

31.  What  must  be  the  value  of  r  in  order  thai 

(1'"'      /7W     ? 

32.  The  ratio  of  the  number  of  combinations  of  2n  things 
in  4:71  to  that  of  the  combinations  of  71  things  in  2/1  is 

{271  +  1){271  +  S)    ,    .    .    .   {4:71  ~  3)  {4:71  -  1) 

1.3.5  .  .  .  .  (2/^-l) 

33.  Show  that  the  sum  of  the  7i\  different  numbers  that 
can  be  formed  by  permuting  any  71  different  digits  is  divis- 
ible by  {71  —  1)  times  the  sum  of  the  digits,  and  that  the 
quotient  is  111  .... 

34.  If  we  define  a  magic  square  as  an  arrange-  6  18 
ment  of  ?i'  numbers  in  a  square  such  that  the  sum 

of  every  line  and  every  column  is  equal  to  the  same  '2'  5  3 
quantity;  show  that  if  one  such  arrangement  is  pos-  004 
sible  with  given  numbers,  then  {niy  are  possible. 


SERIES.  635 

See  margin  for  example  of  square  when  ti  =  3,  and  note  that  we 
leave  out  of  consideration  the  diagonal  lines  of  numbers. 

35.  Given  m  different  letters  and  n  different  numbers, 
find  the  number  of  different  permutations  each  containing  r 
letters  and  s  numbers. 

2t(i,  Given  n  unequal  straight  lines;  how  many  non-identi- 
cal rectangular  parallelopipeds  may  be  formed,  each  of  whose 
edges  must  be  equal  to  some  one  of  these  lines  in  the  two 
cases ;  (1)  When  the  same  line  cannot  be  repeated  in  a 
figure  and  (2)  When  it  can  be  repeated  without  restriction. 

37.  The  same  thing  being  supposed  and  case  (1)  taken; 
how  many  different  parallelopipedons  may  be  built  upon  the 
same  horizontal  plane  as  a  base,  with  their  vertical  faces 
toward  the  four  points  of  the  compass ;  two  figures  being 
regarded  as  different  when  they  cannot  be  brought  into  coin- 
cidence without  turning  them  around  or  over. 

38.  Given  n—1  sets  containing  respectively  2^^,  3«  .  .  .  .  na 
different  things;  show  that  the  number  of  combinations  com- 
prising a  of  the  first  set,  2a  of  the  second,  etc.,  is  ^  ^>1'. 

Series. 

Indeteeminate  Coefficients. 
Develop  the  following  expressions  in  powers  of  x  by  the 


method  of  indeterminate  coefficients: 

I. 

1  +  nx 
1-x' 

2. 

1  +  x 
1  —  nx' 

3- 

1  +mx 
1  +nx' 

4. 

x  +  a 

C  —  X 

5- 

a{a  +  x) 

a'  +  x'' 

6. 

a'  +  x' 
a-\-x' 

•7. 

x'  +  a' 
x"  +  a'* 

8. 

(i^x){i^hxy 

9. 

X 

10. 

x^ 

(1  ^  ex)  (1  - 

CJ 

l  +  ic*" 

1 

12. 

1 

a"  -\-  ax-\'  x^* 

0^  —  ax-^  x^' 

536  SERIES. 


Products  of  Series. 

Form  the  products: 

1.  (1  -  a;  +  x'-  x'+  .  .  .  .){l  +  x+x'+x'+  .  .  .  .)• 

2.  (l+2^+3^'^+4a;^+  .  .  .  .)  {l-^x+^x'-^x^+  ....). 

,  (,j   y\y'  y\      \{^j^y\y\y' jl.       \ 

4.  {i+ax+ av + — )  (i+|+|; +|;+ .  .  .  .). 

5.(i-..+av-....)(i-i+i;-:-;+....). 

6.  (1  +  2a;  +  3a:'' +  4a;' +  .  .  .  .)\ 

7.  (1  -  2a;  +  3a;'' -  4a:' +  .  .  .  ,)\ 

Carry  the  products  as  far  as  x^  and  express  the  7i^^  term 
of  the  product  in  terms  of  n  in  each  case  for  which  you  can 
form  it. 

FiGURATE  Numbers. 

1.  Enumerate  an  incomplete  pile  of  cylindrical  shot  (§  288) 
haying  n  shot  in  its  bottom  row,  and  as  many  in  its  top  row 
as  there  are  rows. 

Show  that  in  this  problem  the  number  n  must  be  odd. 

2.  The  top  and  bottom  rows  of  an  incomplete  pile  of  cylin- 
drical shot,  having  8  rows  in  all,  contain  9  shot  less  than  one 
third  the  pile.     How  many  shot  are  in  the  pile? 

3.  In  an  incomplete  pile  of  63  cylindrical  shot  35  are  in 
the  interior  of  the  pile,  so  as  to  be  completely  surrounded  by 
others,  and  28  form  the  top,  bottom  and  sides.  Describe  the 
pile,  and  show  that  two  piles  may  be  formed  which  fulfil  the 
conditions. 

4.  In  a  triangular  pyramid  of  balls  the  ratio  of  the  whole 
number  of  balls  to  the  number  in  the  bottom  layer  is  14  :  3. 
How  many  balls  form  the  pile? 

5.  In  a  triangular  pyramid  having  n  balls  on  each  edge, 
how  many  balls  form  the  four  faces? 

6.  If  20  balls  in  a  triangular  pyramid  are  completely  sur- 
rounded by  others,  how  many  form  the  entire  pyramid? 


SERIES.  537 

^    7.  A  rectangular  pile  has  15  balls  in  its  top  row  and  its 
lesser  side  has  10  balls.     Enumerate  the  balls  in  the  pile. 

8.  If  one  side  of  the  base  contains  m  balls  and  the  other  n 
(m  >  n),  how  many  balls  will  the  pile  contain;  how  many 
layers,  and  how  many  balls  in  the  top  row? 

9.  If  495  balls  form  a  complete  rectangular  pile,  having 
10  balls  on  one  side  of  the  base,  how  many  will  the  other  side 
comprise? 

10.  How  many  balls  in  a  square  pyramid  having  12  balls 
on  each  side  of  the  base? 

11.  A  rectangular  pile  has  84  shot  in  its  bottom  layer  and 
66  in  the  next  layer.     How  many  in  the  whole  pile? 

Prove: 

12.  1.2  +  2.3  +  3.4+  ....  +7i{n  +  l) 

_  n{n  +  l){n  +  2) 
~  3  • 

13.  17^  +  2(72,-1) +  3 (^-2)+  ,  .  ,  .  +  n[n-{7i-l)] 

__n{n  +  l)  {n  +  2) 
"  3! 

14.  1.2  +  2.4  +  3.6+  ....  +n.2n 

_  n{n  +  l)  (271  +  1) 
""  3 

15.  1  (2  -  ^)  +  2  (4  -  w)  +  3  (6  -  7^)  +  .  .  .  .  +  n' 

_n{n  +  l)  {n  +  2) 
~  3! 

16.  If  we  multiply  the  corresponding  terms  of  the  two 
progressions: 

a,     a  +  Ji,     a  +  2hy  ,  .  ,  ,  a  +  ih, 
i,     1)  —  h,     h  —  2h,  ,  ,  ,  ,  h  —  ill, 

the  s.um  of  the  products  will  be 

(^•+1)  \ai> +'^^^y -i^i±^\ . 

17.  Find  the  sum  of  the  products  when,  in  the  second 
series,  the  C.  D.  is  +  li  instead  ot  —  h. 


538  SERIES. 

Express  the  values  of 

i8.  ai  +  ia^  h)  {h  +  k)  +  (a  +  2A)  {h +  2k) -\-  .  .  .  . 

to  n  terms. 

19.  1.3  +  3.5  +  5.7+  ....  +^(^  +  2). 

20.  1^  +  3  (a  —  3)  +  5  («5  —  6)  +  .  .  .  .to  n  terms. 

21.  l.tt  +  3  (a  +  3)  +  5  (nj  +  6)  +  .  .  .  ,ton  terms. 

22.  Prove  the  equations: 

4  5  6  7 
1.2.3  +  2.3.4  +  3.4.5  +  4.5.6  =  1^1.1 

by  subtracting  from  the  second  member  the  successive  terms 
of  the  first  member,  beginning  with  the  last. 

23.  Generalize  the  preceding  result  by  proving  in  the  same 
way  the  general  equation: 

_  (n  +  1\ 
Note  that  the  first  operation  will  be" to  deduce 

By  means  of  the  preceding  formulae  write,  on  sight,  the 
values  of: 

24.  1.2.3.4  +  2.3.4.5  +  3.4.5.6  +  4.5.6.7 
1.2.3       2.3.4      3.4.5       4.5.6       5.6.7 

^^*  1. 2. 3  "*"  1.2. 3 +  1.2. 3  "^1.2. 3  "^1.2. 3* 

26.  1.2. 3. 4  +  2. 3. 4.5  +  .  .  .  .  +  ?^(^  + 1)  (^+2)  (w+3). 

27.  Show  that  the  sum  of  the  products  of  the  first  n  natu- 

ral  numbers  taken  by  2's  is  ^^ ^- — ^    7^    -^  ^ — \ 

28.  In  the  following  scheme  we  start  with  a  column  of  ar's 
on  the  left,  and  with  the  top  line  a,  ft,  y,  d,  etc.  Then,  each 
number  following,  in  each  column,  is  formed  by  adding  the 
number  above  it  to  the  number  on  the  left  of  the  latter.     It 


8ERIE8. 


539 


is  now  required  to  write  the  general  expression  for  the  ni\i 
number  in  the  2d,  3d,  4th,  and  iih.  columns. 


1 

2 

3 

4 

a 

P 

7 

d 

a 

P^-    a 

r^ 

V  ^ 

^^ 

h  r 

a 

/5  +  2a 

yj 

-215- 

-    a 

6  - 

-2y- 

-    ^ 

a 

y5  — 3a: 

yj 

-3/i- 

-3a 

6- 

-3r- 

-3/i+    a 

a 

y^  +  4a 

r  - 

h4/^H 

h6a 

8  +  ir- 

\-  6/i  +  4a 

29.  A  trader  starts  with  a  capital  of  a  dollars;  he  gains, 
and  adds  to  his  capital,  h  dollars  the  first  year,  and  c  dollars 
more  each  year  than  he  did  the  year  before.  Express  his 
accumulated  capital  at  the  end  of  n  years  in  terms  of  n. 

SuMMATioiT  OF  Series. 
Sum  to  infinity: 

1.  1  +  7^  +  (1  +  2?z)  a:  +  (1  +  Zn)  x^-\-  (1  +  ^n)  2:'+ .... 

2.  1  +  3a;  +  ^x^  +  10^'+  . 


3- 
4. 

5- 
6. 

7. 
8. 


1  +  4a;  +  ^x^  +  16a;'  + 


.  +n'x'"'^-i-  .  .  .  . 


1.3 
1 


2.4 


3.5 
1 


2.5  +  5.8"'"8.11 


+ 


2.4^4.6^6.8^ 


2.3.4 

1 
1.3.5 

1 


+ 


+ 


+ 


3.4.5 
1 

2.4.6 
1 


+ 


+ 


+ 


4.5.6 

1 
3.5.7 

1 


+ 


^    1.4.7    '   2.5.8    '    3.6.9 

10.  l-i^{a+l+a-^)  r  +  {a'+  a  +  1  +  a-^-{-a-y''+ 

11.  {71  +  lyx  +{n  +  2)V  +  {71  +  3)V  +  .  .  .  . 

12.  -  +  -i  +  -:^  +  -4  +  .  .  .  . 


540  LIMITS. 

Sum  to  n  terms: 

13.  a'J^{a  +  iy  +  {a  +  %Y+ 

14.  3  +  5  +  9  +  14  +  .  .  .  .  +  M!i+i). 

z 

15.  3  +  8  +  15  +  24  +  .  ,,,-]- 71  {n  +  2). 

16.  1  +  /^  +  2  (2  +  y?;)  +  3  (3  +  ^)  +  .  .  .  ,  J[-n{n  +  r 
16a,  Show  that  the  series: 

2^3        4^5 
may  be  transformed  into  eitiier  of  the  three  forms: 

Jl_  4.  _1     ■   _1     . 
1.2^3.4^5.6^  *  *  •  • 

.  _  J ^   ___^   _ 

^^  2.3       4.5       6.7       •  •  •  • 

..       11,111 
^^         2  "^1.2.3"^  3.4.5  "^5.6.7"^7.8.9"^**** 

17.  How  do  two  of  the  preceding  results  enable  us  to  sum 

r~2  "^  2~3  "^  3~4  +  *  •  *  *  ^^  wfinitum? 

18.  What  number  is  equal  to  the  continued  product: 

2^. 4i.8T's.  16^^.32^5  ....  ad  infinitum? 

19.  To  what  limit  approaches  the  indefinitely  continued 

product: 

1     ?_     ?_     i_ 
a^.a'^^a^^a»** .  .  .  .  ? 


Limits. 


Find  the  limits  of 


{x  -  by 

x" 


as  X  increases  indefinitely. 


2. 
ax 

ax 
3-  -.- 


LIMITS.  541 

4.  '^ — ~      as  X  approaches  a  indefinitely. 


X  —  a 


X 

a 

x' 

-a' 

x' 

-a' 

a 

X 

X 

a 

12. 


X  —  a 

(x  +  a)''  —  (x  —  ay  .  •    -,  ^   .^  1 

7.  -^^ n^ ^s  X  increases  indefinitely. 

(1  +  axY 

•  (1  +  bxy 

(1  -  axr  ,,  ,,         ,, 

^*  (1  -  bxy 

10.  ^ as  n  increases  indefinitely. 

„  r  +  r  +  3'+....  +  n'    ,,,,      ,. 

in^  4-  2^  +  3"^  +  4^  +  .  .  .  .  +  ^^       ^^ 

13.  The  first  term  of  a  series  is  — ,  the  second  —  tt,  and 

o  6 

each  succeeding  term  one  half  the  sum  of  the  two  which  pre- 
cede it.  To  what  limits  will  the  nth  term  and  the  sum  of 
the  series  approach  as  n  increases  indefinitely? 

14.  Find  the  limit  toward  which  the  nth  term  approaches 
when 

First  term  =  a-{-2b;     second  term  =  a  —  b; 
cc        u     ^  1.  ..         <c      ^  1. 

each,  term  after  the  second  being  half  the  sum  of  the  two 
preceding  terms. 

15.  The  first  term  of  a  series  is  a,  the  second  b,  and  each 
following  one  the  geometrical  mean  of  the  two  preceding  it. 
Show  that,  as  ^increases  indefinitely,  the  ^th  term  approaches 
the  limit  albk 


542  BINOMIAL   THEOREM. 

Binomial  Tlieorem. 

Develop: 
I.  (1  -  x)-\  2.  (1  ~  xy^ 

7.  (a  +  l)K  8-  {a  -  i)K 

g.  (a  +  b)-i.  lo.  (a  -  b)-i. 

13.  (c.-^f.  ^4.  (.•  +  ^f. 

In  the  six  last  developments  arrange  the  result  in  the  form 
A  +  B  [x±l)  +  C{x^  -^^  +etc. 

Develop  as  far  as  x^  and  arrange  in  powers  of  x: 

17.   {1  +  X  +  x')\  18.   (1  -  a:  +  ^y. 

19.  (l  +  x-  ir')-^  20.   (1  -  a;  -  a;')-^ 

21.  Write  the  development  of  (1  —  x)-^  in  such  a  form 
that  the  denominator  of  each  term  shall  be  3 !  and  express  the 
tth  term  as  the  product  of  3  factors. 

22.  Write  the  development  of  (1  —  x)"^""  in  such  a  form 
that  all  the  terms  shall  have  the  common  denominator 
{2n  —  1)!  and  show  that,  putting  for  brevity,  p  =  n-{-i  —  1, 
the  ^th  term  may  be  written  in  the  form 

{2n-l)l  • 

23.  Show  that,  if  n  be  an  odd  number,  and  if  we  put 
p  =  n  -{-  2i  —  2,  the  ith.  term  in  the  development  of  (1  —  a;)~" 
may  be  expressed  in  the  form 

(y  -  1')  (y  -  y)  if  -  5') [f-jn-  2)'] 

3.4.6  .  .  .  .2{n-l) 


BINOMIAL  THEOREM.  543 

24.  Show  that  the  ratio  of  the  ni\\  term  to  the  (71  —  l)th 
in  the  expression  of  (1  —  x)"^  is  2a;. 

Of  what  quantities  are  the  following  series  the  develop- 
ments? 


2    '   2.4       2.4.6   '    •  •  •  • 

25.  1-i-g  ^6.12^6.12.18^  •  •  •  • 

Express  the  general  term  of  the  following  developments: 

29.  (1  +  2^;  +  ^^.  30.   (1  +  2a;  +  a;')-^ 

31.   (1  -  2a;  +  a;y.  32.   (1  -  2:?;  +  a;')-^ 

33.  (1  +  ^  +  ^'^'^  +  ^'  +  •  •  •  •  ^^  infi7iitumY. 

34.  (1  -  a;  -+-  o;^  -  a;'  +  .  .  .  .  "         '^         )^ 

35.  \l-2x  +  2V  -  2V  +...."         "        )-". 

36.  Prove  that 

2-  _  ^^^  2—^+  (1)  2—2  - +  (_  1)-=  1. 

37.  If,  in  the  development  of  (1  +  ^Y  we  call  the  second 
term  a  and  the  third  1),  express  n  and  x  in  terms  of  aj  ana  b. 

Of  what  expressions  are  the  following  series  the  develop- 
ments? 

38.  3"  +  (^')  S™-'  +  (1)  3-^  +....+ 1. 

39.  3- -(f)  3"'-+ (1)3---.... 

,    ,  ,    ,  4    ,4.5    ,    4.5.6    , 

40.  1  +  1+-  +  --  +  ^-^-+.... 

2^t  2?i.dn 

42.  If  ^y  be  the  rth  term  in  the  expansion  of  (1  -+  ^)**+*' 
show  that 

h  +  t,  +  h+ ==(l-^)-<«+«). 


544         LOGARITHMS  AND  LOGABITHMIG  SERIES, 


Exponential  Theorem. 

1.  Find  two  expressions  each  for  the  coefficients  of  x^,  x^, 
and  a;"  in  the  development  of  e'^e^,  and  show  their  identity. 

2.  Develop  e"*^  in  powers  of  x  to  six  terms. 

3.  What  is  the  coefficient  of  x^y^  in  the  development  of 
6^+^?     In  that  of  e^-^? 

4.  Multiply  the  two  developments: 

X^         x^ 
e— r=:l-a;+—  --J+ 

and  show  by  what  relations  among  the  coefficients  the  x>rod- 
uct  reduces  identically  to  unity. 

5.  Show  by  what  relations  the  development  of  e^^  becomes 
identical  with  the  square  of  that  of  e^. 


Logarithms  and  Logarithmic  Series. 

I.  Express  the  logarithm  of  the  continued  product  of  all 
tlio  terms  of  a  geometrical  progression. 

Calling  h  the  arbitrary  base  of  the  system  of  logarithms, 
ve  the  following  equations  so  as  to  express  x  in  terms  of  y: 


2: 

4. 

6. 

8. 

10. 

log  X    =  y,                            3.  log  X     =  ay, 
log  2x  =  y,                          5.  log  7}ix  —  a  -\-  y, 
log  ax  =  my,                        7.  log  x^    =  y, 
logx""  :=  my.                        g,  y           —  V''«'^. 

Eec 

iuce  to  their  simplest  form  the  expressions: 

12. 

Prove  the  identities: 

^log  aj^log  y  ?  £ 

14.     m^^^  =  ^yloemx  -      _^^_ "^ _.  ^losmy^ogn. 


/ 


LOGARITHMS  AND  LOGARITIIMIG  SERIES.         545 

16.  If  a,  h,  and  c  be  the  mth,  ^th  and  ^tli  terms  of  a  geo- 
metrical progression  show  that 

{P  —  q)  log  a-\-{q—  m)  log  h  +  (m  —  p)  log  c  =  0. 

17.  Prove  that  the  value  of  the  expression 

is  independent  of  n  and  equal  to  log  a, 

18.  Prove  the  equation: 

2  log  X  —  log  {x  -\-  a)  —  log  {x  —  a) 
_^.^j       a'         ,   1  ^^  ,1         «"  ,  ) 

-'^^]2a;^  _  ^^+3  '  (2x'-a'y~^  6  (2x'-ay^  '  '  '  '  ]' 

19.  If  a,  J  and  c  are  three  consecutive  numbers  show  that 
21og*-log«-log.=  3Jf]^-+3^J-p^,  +  ....[. 

20.  Prove: 

Nap.  log4  =  1  +  j-|^  +  ^+^+  .... 

21.  If  a,  I,  Cy  d,  etc.y  are  in  geometrical  progression,  then, 

in  order  the  equations 

1111^ 
a-^  =M  =  cp  =  d^=  .  .  .  . 

may  be  satisfied,  the  quantities  m,  n,  p,  q,  etc.,  must  be  in 

arithmetical  progression. 

1  1 

22.  If    2/    =    IQi-iog^,     and     z^K^^-'^^sv^     show    that 


23.  Prove  the  development 

and  by  making  the  development  in  another  form  and  com- 
paring the  coefficients  of  x^  prove  the  identity 

2"-g_o.-.      «-3g„_,   I    (w-4)(«-5) 

the  series  terminating  with  tlie  last  exponent  which  is  not 
negative.  


HINTS  ON  A  COURSE  IN  ADVANCED  ALGEBRA. 


For  the  benefit  of  students  wlio  may  contemplate  a  course  of  reading 
in  the  various  branches  of  Advanced  Algebra,  the  following  list  of  sub- 
jects and  books  has  been  prepared.  As  a  general  rule,  the  most  extended 
and  thorough  treatises  are  in  the  German  Language,  while  the  French 
works  are  noted  for  elegance  and  simplicity  in  treatment. 

To  pursue  any  of  these  subjects  to  advantage,  the  student  should  be 
familiar  with  the  Differential  Calculus. 

I.  THE  GENERAL  THEORY  OF  EQUATIONS.— In  English,  Tod- 
hunter's  is  the  work  most  read. 

Serret,  Algehre  Superieurey  2  vols.,  8vo,  is  the  standard  French  work, 
covering  all  the  collateral  subjects. 

oToT?T»A^^  Theorie  des  Substitutions  et  des  Equations  Algefyriques,  1  vol.,  4to, 
s  the  largest  and  most  exhaustive  treatise,  but  is  too  abstruse  for 
my  but  experts. 

_-.  -.i^PERMINANTS.— Baltzer,  Theorie  der  Beterminanten,  is  the 
standard  treatise.  There  is  a  French  but  no  English  translation. 
A  recent  English  work  is  Robert  F.  Scott,  The  Theory  of  Deter- 
minants and  their  Applications  in  Analysis  and  Geometry. 

III.  THE  MODERN  HIGHER  ALGEBRA,  resting  on  the  theories  of 

Invariants  and  Covariants. 
Salmon,   Lessons  introductory  to  the  Modern  Higher  Algebra^  is  the 

standard  English  work,  and  is  especially  adapted  for  instruction. 
CLEB8CH,  Theorie  der  lindren  Algebraischen  Formen^  is  more  exhaustive 

in  its  special  branch  and  requires  more  familiarity  with  advanced 

systems  of  notation. 

IV.  THE  THEOKi.  OF  NUMBERS.      There  is  no  recent  treatise  in 

English.  Gauss,  Disquisitiones  Arithmeticm,  and  Legendre, 
Theorie  des  Norribres,  are  the  old  standards,  but  the  latter  is  rare 
and  costly.  Lejeune  Dirichlet,  Vorlesungen  uber  Zahlentheorie, 
is  a  good  German  Work.  There  is  also  a  chapter  on  the  subject  in 
Serret,  Algehre  Super ieure. 

V.  SERIES. — This  subject  belongs  for  the  most  part  to  the  Calculus,  but 

Catalan,  Traite  eUmentaire  des  Series,  is  a  very  convenient  little 
French  work  on  those  Series  which  can  be  treated  by  Elementary 
Algebra. 

VI.  Q  JATERNIONS.— Tait,  Elementary  Treatise  on  Quaternions^  is 

prepared  especially  for  students,  and  contains  mfiny  exercises.  The 
original  works  of  Hamilton,  Lectures  on  Quaternions  and  Elements 
of  Quaternions,  are  more  extended,  and  the  latter  will  be  found 
valuable  for  both  reading  and  reference. 


UNIVERSITY  OB  CALIFORNIA  LIBRARY 
DUE  on  the  last  date  stamped  below. 


OCT  21   1947 


0cll4'48J§ 


LD  21-lOOw 


.l2,'46(A2012sl6)4120 


Wl^t?5'?  ' 


Xf^*  » 


4 


UNIVERSITY  OF  CAUFORNIA  UBRARY 


lt-M.^1^ 


